Definite Integrals

Let $I=\int _{\pi /6}^{\pi /2}\frac{\cos ecx\cot x}{1+\cos e{{c}^{2}}x}dx=\int _{\pi /6}^{\pi /2}\frac{\cos x}{1+{{\sin }^{2}}x}dx$ Let $\sin x=t$ $\Rightarrow \cos xdx=dt$ Also, when...

Let $I=\int _{1}^{2}\frac{1}{x{{(1+{{\log }_{e}}x)}^{2}}}dx$ Let $1+{{\log }_{e}}x=t$ $\Rightarrow \frac{1}{x}dx=dt$ Also, when $x=1,t=1$and when $x=2$ $t=1+{{\log }_{e}}2$ Hence, $I=\int... read more Let$I=\int _{0}^{{{\left( \frac{\pi }{2} \right)}^{\frac{1}{3}}}}{{x}^{2}}\sin \left( {{x}^{3}} \right)dx$Let${{x}^{3}}=t\Rightarrow 3{{x}^{2}}=dt$Also, when$x=0,t=0$and when$t=\frac{\pi...

Let $I=\int _{\pi /4}^{\pi /2}\frac{\cos x}{{{\left( \cos \left( \frac{x}{2} \right)+\sin \left( \frac{x}{2} \right) \right)}^{3}}}dx$ $\cos x={{\cos }^{2}}\left( \frac{x}{2} \right)-{{\sin... read more Let$I=\int _{2}^{3}\frac{2-x}{\sqrt{5x-6-{{x}^{2}}}}dx$Let$2-x=\frac{ad}{dx}\left( 5x-6-{{x}^{2}} \right)+b=-2ax+5a+b$Hence$-2a=-1$and$5a+b=2$Solving these equations, We get and... read more Let$I=\int _{0}^{\pi /2}\left( \sqrt{\tan x}+\sqrt{\cos tx} \right)dx=\int _{0}^{\pi /2}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx$Let$\sin x-\cos x=t\Rightarrow \left( \cos x+\sin x...

Let $I=\int _{0}^{1}\frac{1-{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}dx$ Let $x=\tan t$ $\Rightarrow dx={{\sec }^{2}}tdt$ Also, when $x=0,t=0$and when $x=1,t=\frac{\pi }{4}$ Hence, $I'=\int _{0}^{\pi... read more Let$I=\int _{0}^{1}{{x}^{3}}\sqrt{1+3{{x}^{4}}}dx$Let$1+3{{x}^{4}}=t\Rightarrow 12{{x}^{3}}dx=dt$Also, when$x=0,t=1$and when$x=1,t=4I=\frac{1}{12}\int _{1}^{4}\sqrt{t}dt$... read more Let$I=\int _{0}^{9}\frac{1}{1+\sqrt{x}}$Let$\sqrt{x}=u\Rightarrow \frac{1}{2\sqrt{x}}=dx=du$Or$dx=2udu=\frac{1}{2u}dx$Also, when$x=0,u=0$and$x=9,u=3$Hence,$I=\int...

Let $I=\int _{0}^{a}{{\sin }^{-1}}\sqrt{\frac{x}{a+x}}dx$ Let $x=a{{\tan }^{2}}y$ $\Rightarrow dx=2a\tan y{{\sec }^{2}}ydy$ Also, when $x=0,y=0$ And when $x=a$ $y=\frac{\pi }{4}$ Hence, $I=\int... read more Let$I=\int _{0}^{1}{{\sin }^{-1}}\sqrt{x}dx$Let$\sqrt{x}=1\Rightarrow \frac{1}{2\sqrt{x}}=dx=dt$Or$dx=2tdt$When$x=0,t=0$And when$x=1,t=1$Hence,$I=2\int _{0}^{1}t{{\sin }^{-1}}tdt$... read more Let$I=\int _{0}^{1}x{{(a{{\tan }^{-1}}x)}^{2}}dx$Using integration by parts, we get$I=\frac{{{\left( {{\tan }^{-1}}x \right)}^{2}}{{x}^{2}}}{2}|_{0}^{1}-\int _{0}^{1}\frac{2{{\tan...

Let $I=\int _{0}^{1}{{\left( {{\cos }^{-1}}x \right)}^{2}}dx$ Let $x=\cos t\Rightarrow dx=-\sin tdt$ Also, when $x=0,t=\frac{\pi }{2}$ And when $x=1,t=0$ Hence, $I=-\int _{\pi /2}^{0}{{t}^{2}}\sin... read more Let $I=\int _{\pi /3}^{\pi /2}\frac{\sqrt{1+\cos x}}{(1-\cos x)\frac{5}{2}}dx$ Using$1+\cos x=2\cos {{x}^{2}}\left( \frac{x}{2} \right)1-\cos x=2{{\sin }^{2}}\left( \frac{x}{2} \right)$we get... read more Let$I=\int _{0}^{\pi /2}\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx$Using$\sin 2x=2\sin x\cos x$, we get$=2\int _{0}^{\pi /2}\frac{\tan x{{\sec }^{2}}x}{({{\tan }^{4}}x+1)}dxI=\int...

Let $I=\int _{0}^{\pi /2}\frac{\sin x\cos x}{{{\cos }^{2}}x+3\cos x+2}dx$ Let $\cos x=t$ $\Rightarrow -\sin xdx=dt$ Also, when $x=0,t=1$ And when $t=0$ $x=\frac{\pi }{2}$ Hence, $I=-\int... read more Let$I=\int _{0}^{\pi /4}\frac{{{\tan }^{3}}x}{1+\cos 2x}dx$Using$1+\cos 2x=2{{\cos }^{2}}x$, we get$I=\frac{1}{2}\int _{0}^{\pi /4}{{\tan }^{3}}x{{\sec }^{2}}xdx$Let$\tan x=t\Rightarrow...

Let $I=\int _{0}^{\pi }\frac{1}{3+\cos x+2\sin x}dx$ Using $\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ $\cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2}... read more$f(x)$is continuous in$\left[ a,b \right]\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where$h=(b-a)/n$Here$h=\left( b-a \right)/n\int...

$f\left( x \right)$is continuous in $\left[ 1,3 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{1}^{3}\left(... read more$f\left( x \right)$is continuous in$\left[ 0,2 \right]\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where$h=(b-a)/n$Here$h=2/n\int _{0}^{2}\left(...

Since it is modulus function so we need to break the function and then solve it $f\left( x \right)=\int _{0}^{1/3}\left( 1-3x \right)dx+\int _{1/3}^{1}(3x-1)dx$it is continuous in $\left[ 0,1... read more$f\left( x \right)$is continuous in$\left[ 0,3 \right]\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where$h=(b-a)/n$Here$h=3/n\int _{0}^{3}\left(...

$f\left( x \right)$is continuous in $\left[ 0,2 \right]$ $\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$ Here $h=2/n$ $\int _{0}^{2}\left(... read more$f(x)$is continuous in$\left[ 2,4 \right]\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where$h=(b-a)/n$Here$h=3/n$$\int... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx Using \sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)} And \cos x=\frac{1-{{\tan... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{\cos x+2\sin x}dx Using \sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)} And \cos x=\frac{1-{{\tan... read more Let I=\int{_{0}^{\pi }}\frac{1}{5+4\cos x}dx Using, we get \cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)} I=\int{_{0}^{\pi... read more I=\int{_{0}^{\pi }}\frac{1}{6-\cos x}dx Using, we get \cos x=\frac{1-{{\tan }^{2}}\left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)} I=\int{_{0}^{\pi }}\frac{1-{{\tan... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{5+4\sin x}dx \sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{4+9{{\cos }^{2}}x}dx Dividing by {{\cos }^{2}}x in numerator and denominator, we get I=\int{_{0}^{\frac{\pi }{2}}}\frac{{{\sec }^{2}}x}{2{{\sec... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\frac{1}{1+{{\cos }^{2}}x}dx Dividing by {{\cos }^{2}}xin numerator and denominator, we get I=\int{_{0}^{\frac{\pi }{2}}}\frac{{{\sec... read more Let I=\int{_{0}^{\frac{\pi }{2}}\frac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}dx} Dividing by {{\cos }^{2}}xin numerator and denominator, we get I=\int{_{0}^{\frac{\pi... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\sin x}dx Using and {{\sin }^{2}}\frac{x}{2}+\cos \frac{x}{2}=1 \sin x=2\sin \frac{x}{2}\cos \frac{x}{2} I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{{{\left(... read more Let I=\int{_{0}^{\frac{\pi }{2}}}\sqrt{1+\cos x}dx Using, we get 1+\cos x=2{{\cos }^{2}}\frac{x}{2} I=\sqrt{2\sin }\left( \frac{x}{2} \right)\left| _{0}^{\frac{\pi }{2}} \right.=2 read more Let I=\int{_{0}^{1}}{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)dx Let \int{\left( x \right)}={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) Let x=\tan \theta \Rightarrow \theta... read more Let I=\int{_{0}^{2}x\sqrt{2-xdx}} Using the property that, we get \int{_{a}^{b}f(x)dx=\int{_{a}^{b}f(a+b-x)dx}} I=\int{_{0}^{2}(2-x)\sqrt{xdx}}... read more Let I=\int{_{0}^{1}\frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx} Let {{a}^{2}}+{{x}^{2}}={{t}^{2}} \Rightarrow xdx=tdt. Also when x=0,t=a And when x=a, t=\sqrt{2a} Hence ,... read more Let I=\int{_{0}^{a}\frac{{{x}^{4}}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} Let x=a sin t a cos t dt=dx also, when x=0,t=0 and when x=a , t=\frac{\pi }{2} Hence, I=\int{_{0}^{\frac{\pi... read more Let I=\int{_{0}^{\sqrt{2}}\sqrt{2-{{x}^{2}}}dx} Consider, I=\int{_{0}^{\sqrt{2}}\sqrt{2-{{x}^{2}}}dx} Let x=a sin t \Rightarrow a\cos ttdt=dx Also, when x=0,t=0 and when x=a, t=\frac{\pi... read more Let I=\int{_{0}^{a}\sqrt{{{a}^{2}}-{{x}^{2}}}dx} Let x=a sin t \Rightarrow a\cos tdt=dx Also, x=0,t=0 And When x=a,t=\frac{\pi }{2} Hence, I=\int{_{0}^{\frac{\pi... read more I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x\cos x}{1+{{\sin }^{4}}x}dx} Let {{\sin }^{2}}x=t \Rightarrow 2\sin x\cos x=dt. Also, When x=0,t=0 And When x=\frac{\pi }{2},t=1. Hence,... read more Let I=\int{_{0}^{\frac{\pi }{2}}\sqrt{\sin x{{\cos }^{5}}xdx}} Let \sin x=t \Rightarrow \cos xdx=dt. Also, When x=0,t=0 And When x=\frac{\pi }{2},t=1. Consider {{\cos }^{5}}x={{\cos... read more Let I=\int{_{0}^{\frac{\pi }{2}}\frac{\sin x}{\sqrt{1+\cos x}}dx} Let 1+cosx=t \Rightarrow -\sin xdx=dt. Also, when x=0,t=2 And When x=\frac{\pi }{2},t=1 Hence,... read more Let I=\int{_{0}^{1}}\sqrt{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx} Let {{\tan }^{-1}}x=t \Rightarrow \frac{1}{1+{{x}^{2}}}dx=dt When x=1, t=\frac{\pi }{4} Hence, I=\int{_{0}^{\frac{\pi... read more f\left( x \right)is continuous in \left[ 0,2 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n Here h=2/n \int _{0}^{2}\left(... read more f\left( x \right)is continuous in \left[ 1,3 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n \int _{1}^{3}\left(... read more f\left( x \right)is continuous in \left[ 1,3 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n Here h=3/n \[\int... read more f\left( x \right)is continuous in \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n Here h=3/n \[\int... read more f\left( x \right)is continuous in \left[ 2,5 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=\left( b-a \right)/n Here h=3/n \int... read more f(x)is continuous in \left[ 2,5 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+r-h), where h=(b-a)/n Here h=3/n \int... read more f(x)is continuous in \left[ 0,3 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=b-a/n Here h=3/n \[\int... read more f\left( x \right)is continuous in \left[ 1,3 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=\left( b-a \right)/n Here h=2/n \int... read more f(x)is continuous in \left[ 1,2 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n Here h=1/n \int... read more f(x)is continuous in \left[ 0,2 \right] \int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh), where h=(b-a)/n Here h=2n \int _{0}^{2}\left( x+4... read more y=\int _{0}^{4}\left| x \right|+\left| x-2 \right|+\left| x-4 \right|dx y=\int _{0}^{2}\left| x \right|+\left| x-2 \right|+\left| x-4 \right|dx+\int _{2}^{4}\left| x \right|+\left| x-2... read more y=\int _{0}^{4}f(x)dx y=\int _{0}^{2}f(x)dx+\int _{2}^{4}f(x)dx y=\int _{0}^{2}3{{x}^{2}}+4dx+\int _{2}^{4}9x-2dx y=\left( {{x}^{3}}+4x \right)_{0}^{2}+\left( \frac{9{{x}^{2}}}{2}-2x... read more y=\int _{1}^{3}f(x)dx y=\int _{1}^{2}f(x)dx+\int _{2}^{3}f(x)dx y=\int _{1}^{2}2x+1dx+\int _{2}^{3}{{x}^{2}}+1dx y=\left( {{x}^{2}}+x \right)_{1}^{2}+\left( \frac{{{x}^{3}}}{3}+x... read more We know that \left| \sin x \right|=-\sin xin \left[ -\pi /4,0 \right) \left| \sin x \right|=\sin xin \left[ 0,\pi /4 \right] y=\int _{-\pi /4}^{0}\left| \sin x \right|dx+\int _{0}^{\pi... read more We know that \left| \cos x \right|=\cos xin \left[ 0,\pi /2 \right) \left| \cos x \right|=-\cos xin \left[ \pi /2,3\pi /2 \right) \left| \cos x \right|=\cos xin \left[ 3\pi /2,2\pi ... read more We know that \left| x-5 \right|=-(x-5)in \left[ 0,5 \right) \left| x-5 \right|=(x-5)in \left[ 5,8 \right] y=\int _{0}^{5}\left| x-5 \right|dx+\int _{5}^{8}\left| x-5 \right|dx y=\int... read more We know that \left| x+1 \right|=-(x+1)in \left[ -2,-1 \right) \left| x+1 \right|=(x+1)in \left[ -1,2 \right] y=\int _{-2}^{-1}\left| x+1 \right|dx+\int _{-1}^{2}\left| x+1 \right|dx... read more We know that \left| x \right|=-xin \left[ -1,0 \right) \left| x \right|=xin \left[ 0,1 \right] y=\int _{-1}^{0}{{e}^{\left| x \right|}}dx+\int _{0}^{1}{{e}^{\left| x \right|}}dx y=\int... read more y=\int _{-\pi }^{\pi }{{x}^{12}}{{\sin }^{9}}xdx......(1) Use king theorem of definite integral \int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt y=\int _{-\pi }^{\pi }{{\left( \pi -\pi -x... read more y=\int _{-\pi }^{\pi }{{\sin }^{75}}x+{{x}^{125}}dx.....(1) Use king theorem of definite integral \int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt y=\int _{-\pi }^{\pi }{{\sin }^{75}}\left( \pi -\pi... read more y=\int _{-a}^{a}{{x}^{3}}\sqrt{{{a}^{2}}-{{x}^{2}}dx}.....(1) Use king theorem of definite integral \int _{a}^{b}f(t)dt=\int _{a}^{b}f(a+b-t)dt y=\int... read more Let x=\tan t \Rightarrow dx={{\sec }^{2}}tdt At x=0,t=0 At x=1,t=\pi /4 \[y=\int _{0}^{\pi /4}\frac{\log \left( 1+\tan t \right)}{1+{{\tan }^{2}}t}{{\sec }^{2}}tdt$ $y=\int _{0}^{\pi... read more Use integration by parts \int I\times IIdt=I\int IIdt-\int \frac{d}{dt}I\left( \int IIdt \right)dt y=\log x\int \frac{\sqrt{1}}{\sqrt{1-{{x}^{2}}}}dx-\int \frac{d}{dx}\log x\left( \int... read more Let, x=\sin t \Rightarrow dx=\cos tdt At x=0,t=0 At x=1,t=\pi /2 y=\int _{0}^{\pi /2}\frac{{{\sin }^{-1}}\sin t}{\sin t}\cos tdt y=\int _{0}^{\pi /2}\frac{t\cos t}{\sin t}dt y=\int... read more Use integration by parts \int I\times IIdx=I\int IIdx-\int \frac{d}{dx}I\left( \int IIdx \right)dx y=x\int \cot xdx-\int \frac{d}{dx}x\left( \int \cot xdx \right)dx y=\left( x\log \sin x... read more Use king theorem of definite integral \int _{a}^{b}f(x)=\int _{a}^{b}f(a+b-x)dx y=\int _{1}^{4}\frac{\sqrt{4+1-x}}{\sqrt{4+1-x}+\sqrt{x}}dx y=\int... read more Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{a/4}^{3a/4}\frac{\sqrt{\frac{3a}{4}+\frac{a}{4}-x}}{\sqrt{\frac{3a}{4}+\frac{a}{4}-x}+\sqrt{x}}dx... read more Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{\pi /4}^{3\pi /4}\frac{\left( \frac{3\pi }{4}+\frac{\pi }{4}-x \right)}{1+\sin \left( \frac{3\pi... read more y=\frac{1}{2}\int _{\pi /4}^{3\pi /4}{{\sec }^{2}}\frac{x}{2}dx y=\frac{1}{2}\left( \frac{\frac{\tan x}{2}}{\frac{1}{2}} \right)_{\pi /4}^{3\pi /4} y=\tan \frac{3\pi }{8}-\tan \frac{\pi }{8}... read more Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{\pi /6}^{\pi /3}\frac{\sqrt{\cos \left( \frac{\pi }{3}+\frac{\pi }{6}-x \right)}}{\left( \sqrt{\sin... read more Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{\pi /8}^{3\pi /8}\frac{\cos \left( \frac{3\pi }{8}+\frac{\pi }{8}-x \right)}{\sin \left( \frac{3\pi... read more \[y=\int _{0}^{\pi /2}\log \frac{1}{\sin x\cos x}dx$ $y=-\left( \int _{0}^{\pi /2}\log \sin xdx+\int _{0}^{\pi /2}\log \cos xdx \right)$ Let,$I=\int _{0}^{\pi /2}\log \sin xdx......(1)$Use king... read more Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\log (1+\cos (\pi -x))dx$ $y=\int _{0}^{\pi }\log (1-\cos x)dx........(2)$ Adding... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi }\left( \pi -x \right)\log \sin \left( \pi -x \right)dxy=\int _{0}^{\pi }\pi \log \sin...

$y=\int _{0}^{\pi /2}\log 2+\log \sin x+\log \cos xdx$ Let, $I=\int _{0}^{\pi /2}\log \sin xdx.....(1)$ Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)$ $I=\int... read more ### Let$\Rightarrow \cos x-\sin xdx=dt$At$x=0,t=1$At$x=\pi /2,t=1y=\int _{1}^{1}-\log tdt$We know that when upper and lower limit in definite integral is equal then value of integration is zero.... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi }{{\sin }^{2m}}\left( \pi -x \right){{\cos }^{2m+1}}\left( \pi -x \right)dxy=-\int _{0}^{\pi...

Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }{{\sin }^{2}}\left( \pi -x \right){{\cos }^{3}}\left( \pi -x \right)dx$ $y=-\int _{0}^{\pi... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx.......(2)$Adding equation (1) and (2)$2y=\int...

### Let

$\Rightarrow dx=a\cos tdt$ At $x=0,t=0$ At $x=a,t=0$ $y=\int _{0}^{\pi /2}\frac{a\cos t}{a\sin t+\sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}t}}dt$ $y=\int _{0}^{\pi /2}\frac{\cos t}{\sin t+\cos t}dt$...

Let, $x=\tan t$ $\Rightarrow dx={{\sec }^{2}}tdt$ At $x=0,t=0$ At $x=\infty ,t=\pi /2$ $y=\int _{0}^{\pi /2}\frac{\tan t}{(1+\tan t)(1+{{\tan }^{2}}t)}{{\sec }^{2}}tdt$ $y=\int _{0}^{\pi... read more$y=\int _{0}^{\pi /2}\log \frac{{{\cos }^{2}}x}{2\sin x\cos x}dxy=\int _{0}^{\pi /2}\log \left( \frac{1}{2}\cot x \right)dx.....(1)$Use king theorem of definite integral$\int...

Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)}{1+{{\sin }^{2}}\left( \pi -x \right)}dx$ $y=\int _{0}^{\pi... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi }\frac{\left( \pi -x \right)\sin \left( \pi -x \right)}{1+\sin \left( \pi -x \right)}dxy=\int...

$y=\int _{0}^{\pi }\frac{x\sin x}{1+{{\cos }^{2}}x}dx......(1)$ Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)\sin... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi /2}\frac{{{\cos }^{2}}\left( \frac{\pi }{2}-x \right)}{\sin \left( \frac{\pi }{2}-x \right)+\cos...

Use king theorem of definite  integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi }\frac{\left( \pi -x \right)\tan \left( \pi -x \right)}{\sec \left( \pi -x \right)\cos ec\left(... read more$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{2}(2-x)\sqrt{x}dxy=\int _{0}^{2}2{{x}^{\frac{1}{2}}}-{{x}^{\frac{1}{2}}}dxy=\left(...

Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{1}(1-x){{x}^{5}}dx$ $y=\int _{0}^{1}{{x}^{5}}-{{x}^{6}}dx$ $y=\left(... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi /2}\frac{\sin \left( \frac{\pi }{2}-x \right)-\cos \left( \frac{\pi }{2}-x \right)}{1+\sin...

$y=\int _{0}^{\pi /2}\frac{\sqrt{\sin x}}{\left( \sqrt{\sin x}+\sqrt{\cos x} \right)}dx......(1)$ Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int... read more y=\int _{0}^{\pi /2}\frac{\sqrt{\cos x}}{\left( \sqrt{\sin x}+\sqrt{\cos x} \right)}dx.....(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi... read more Q15. y=\int _{0}^{\pi /2}\frac{1+}{1+\sqrt{\frac{\sin x}{\cos x}}}dx y=\int _{0}^{\pi /2}\frac{\sqrt{\cos x}}{\left( \sqrt{\sin x}+\sqrt{\cos x} \right)}dx.....(1) Use king theorem of definite... read more y=\int _{0}^{\pi /2}\frac{{{\sin }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}dx......(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi... read more y=\int _{0}^{\pi /2}\frac{{{\cos }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}dx......(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi... read more y=\int _{0}^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx....(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx \[y=\int _{0}^{\pi /2}\frac{\sin \left( \frac{\pi... read more y=\int _{0}^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx......(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi /2}\frac{\cos \left( \frac{\pi... read more y=\int _{0}^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx....(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi /2}\frac{\cos \left( \frac{\pi... read more y=\int _{0}^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx....(1) Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi /2}\frac{\sin \left( \frac{\pi... read more Use king theorem of definite integral \int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx y=\int _{0}^{\pi /2}\frac{{{\sin }^{n}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{n}}\left( \frac{\pi }{2}-x... read more Use king theorem of definite integral \[\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ \[y=\int _{0}^{\pi /2}\frac{{{\sin }^{\frac{3}{2}}}\left( \frac{\pi }{2}-x \right)}{{{\sin...

Use king theorem definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi /2}\frac{{{\cos }^{\frac{1}{4}}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{\frac{1}{4}}}\left(... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi /2}\frac{{{\cos }^{4}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{4}}\left( \frac{\pi }{2}-x...

Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi /2}\frac{{{\cos }^{4}}\left( \frac{\pi }{4}-x \right)}{{{\sin }^{4}}\left( \frac{\pi }{2}... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{0}^{\pi /2}\frac{{{\sin }^{7}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{7}}\left( \frac{\pi }{2}-x...

Use king theorem of definite integral $\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$ $y=\int _{0}^{\pi /2}\frac{{{\cos }^{3}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{3}}\left( \frac{\pi }{2}-x... read more Use king theorem of definite integral$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dxy=\int _{b}^{\pi /2}\frac{{{\sin }^{3}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{3}}\left( \frac{\pi }{2}-x...

$\int{_{\frac{1}{4}}^{\frac{1}{2}}\frac{dx}{\sqrt{x-{{x}^{2}}}}=\int{_{\frac{1}{4}}^{\frac{1}{2}}\frac{1}{\sqrt{\frac{1}{4}-{{\left( x-\frac{1}{2} \right)}^{2}}}}}}$ Substitute: $2x-1=u$ $\therefore... read more ### Assume that$\int{_{0}^{2}\frac{1}{-{{x}^{2}}+ax+{{a}^{2}}}dx=-\int{_{0}^{2}\frac{1}{{{x}^{2}}-ax-{{a}^{2}}}dx}}\int{_{0}^{2}\frac{4}{(2x+(-\sqrt{5}-1)a)(2x+\sqrt{5}-1)a)}dx}$... read more ### By reduction formula:$\int{_{0}^{\frac{\pi }{2}}{{\cos }^{4}}xdx=\frac{{{\cos }^{3}}(x)\sin (x)}{4}+\frac{3}{8}\left[ \frac{\sin 2x}{2}+x \right]}=\frac{{{\cos }^{3}}\left( \frac{\pi }{2} \right)\sin \left( \frac{\pi...

$\int{_{\frac{\pi }{3}}^{\frac{\pi }{4}}\left( \frac{{{\tan }^{2}}x+1}{\tan x} \right)dx=\int{_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{{{\sec }^{2}}x({{\tan }^{2}}x+1)}{{{\tan }^{2}}x}dx}}$...

$=\left[ \frac{a}{2}\left( \frac{\sin \pi }{2}+\frac{\pi }{2} \right)+\frac{b}{2}\left( \frac{\pi }{2}-\frac{\sin \pi }{2} \right)-\frac{a}{2}\left( \frac{\sin 0}{2}+0 \right)-\frac{b}{2}\left(... read more ### . Substitute$4x+1\sqrt{7}=u\therefore dx=\frac{\sqrt{7}}{4}du$Now solving:$\int{\left( \frac{1}{{{u}^{2}}}+1 \right)du={{\tan }^{-1}}u}\frac{2}{\sqrt{7}}\int{\left( \frac{1}{{{u}^{2}}+1}...

Substitute: $x+2=u$ $\therefore dx=du$ $=\int{\frac{du}{\sqrt{{{u}^{2}}-1}}}$ $=\log (\sqrt{{{u}^{2}}-1}+u$ Undo substitution: $x+2=u$ $\therefore \int{_{0}^{4}\frac{dx}{\sqrt{{{x}^{2}}+4x+3}}=\log... read more Substitute:$\frac{x+1}{\sqrt{2}}=u\therefore dx=\sqrt{2}du=\int{\frac{\sqrt{2}du}{\sqrt{2{{u}^{2}}+2}}}=\log (\sqrt{{{u}^{2}}+1}+u$Undo substitution:$u=\frac{x+1}{\sqrt{2}}\therefore...

$=\int{_{3}^{4}\frac{dx}{4(x-4)}=\int{_{3}^{4}\frac{1}{4(x+2)}dx}}$ $=\frac{1}{4}\log (x-2)-\frac{1}{4}\log (x+2)$ $=\frac{1}{4}\log 3-\frac{1}{4}\log 1-\frac{1}{4}\log 6+\frac{1}{4}\log 5$...

$=\frac{3}{2}\log x-\log (x+2)$ $=\frac{1}{2}(\log 2+\log 3)$

$\int{_{1}^{2}\frac{1}{(x+1)}dx=\int{_{1}^{2}\frac{1}{(x+2)}dx}}$ $=[\log (x+1)-\log (x+2)]_{1}^{2}$ $=2\log 3-3\log 2$

$=\int{_{0}^{2}}\frac{{{x}^{4}}+1}{{{x}^{2}}+1}dx=\int{_{0}^{2}\frac{2}{{{x}^{2}}+1}dx}$ $=\int{_{0}^{2}}\frac{({{x}^{2}}-1)({{x}^{2}}+1)}{{{x}^{2}}+1}dx+\int{_{0}^{2}\frac{2}{{{x}^{2}}+1}dx}$...

$={{2}^{\frac{3}{2}}}\left( \sin \left( \frac{\pi }{4} \right)-0 \right)$ $=\frac{2\sqrt{2}}{\sqrt{2}}$ $=2$

$={{2}^{\frac{2}{5}}}\left( 0-\sin \left( -\frac{\pi }{4} \right) \right)$ $=\frac{2\sqrt{2}}{\sqrt{2}}$ $=2$

$=\frac{1}{2}\left[ \frac{-\cos 5x}{5}+\cos x \right]$ $=\frac{1}{2}\left[ -\frac{\cos (5\pi )}{5}+\cos (\pi ) \right]-\frac{1}{2}\left[ -\frac{\cos (0)}{5}+\cos (0) \right]$ $=\frac{1}{2}\left[... read more$=\frac{1}{2}\left[ \frac{\sin 3x}{3}+\sin x \right]=\frac{1}{2}\left[ \frac{\sin \left( \frac{\pi }{3} \right)}{3}+\sin \left( \frac{\pi }{6} \right) \right]-0=\frac{1}{2}\left[...

### .

$=\frac{1}{2}\int{_{0}^{\frac{\pi }{4}}(cox-\cos 5x)dx}$ $=\frac{1}{2}\left[ \sin x-\frac{\sin 5x}{5} \right]$ $=\frac{1}{2}\left[ \sin \left( \frac{\pi }{4} \right)-\frac{\sin \left( \frac{5\pi... read more$\int{_{\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{2{{\sin }^{2}}x}=\int{_{\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}\frac{dx}{2}\cos e{{c}^{2}}xdx}}\int{_{\frac{\pi }{2}}^{\frac{\pi...

$\int{_{0}^{\frac{\pi }{4}}\frac{dx}{1+\cos 2x}=\int{_{0}^{\frac{\pi }{4}}\frac{1}{2}{{\sec }^{2}}xdx}}$ $\int{_{0}^{\frac{\pi }{4}}\frac{1}{2}{{\sec }^{2}}xdx=\frac{1}{2}[\tan x]}$...

Let $u=\left( \tan \left( \frac{x}{2} \right)+1 \right)$ $dx=\frac{2}{{{\sec }^{2}}\left( \frac{x}{2} \right)}du$ $=-\frac{2}{u}$ $=-\frac{2}{\tan \left( \frac{x}{2} \right)+1}$ $=2$

$\int{_{0}^{\frac{\pi }{4}}(\cos x-\sin x)dx}$ $=[\sin x+\cos x]$ $=\left[ \cos \left( \frac{\pi }{4} \right)+\sin \left( \frac{\pi }{4} \right)-\cos 0-\sin 0 \right]$ $=\left[... read more ### .$=\sqrt{2}\left| \sin x \right|\sqrt{2}\left| \sin \left( \frac{\pi }{4} \right)-\sin 0 \right|\sqrt{2}\left[ \frac{1}{\sqrt{2}} \right]=1$read more$=\int{_{\frac{\pi }{4}}^{\frac{\pi }{4}}(\cos e{{c}^{2}}(x)-3\cos ec(x)\cot (x)dx}$read more$\frac{1}{4}\int{_{0}^{\frac{\pi }{2}}(3\sin x-\sin 3x)dx=\frac{1}{4}\left[ -3\cos x+\frac{\cos 3x}{3} \right]}=\frac{1}{4}\left[ -3\cos \left( \frac{\pi }{2} \right)+\frac{\cos \left( \frac{3\pi...

$\frac{1}{4}\int{_{0}^{\frac{\pi }{3}}(3\cos x-\cos 3x)dx=\frac{1}{4}\left[ 3\sin x+\frac{\sin 3x}{3} \right]}$ $\frac{1}{4}\left[ 3\sin \left( \frac{\pi }{3} \right)+\frac{\sin \pi }{3}... read more$=-\log \left| \cos ec\left( \frac{\pi }{4} \right)+\cot \left( \frac{\pi }{4} \right) \right|+\log \left| \cos ec\left( \frac{\pi }{6} \right)+\cot \left( \frac{\pi }{6} \right) \right|=-\log...

$=\log \left| \sec \left( \frac{\pi }{3} \right) \right|-in\left| \cos 0 \right|$ $=\log \left| 2 \right|-\log \left| 1 \right|$ $=\log 2$

$\int{_{0}^{\frac{\pi }{4}}({{\sec }^{2}}x-1)dx=\int{_{0}^{\frac{\pi }{4}}\left[ \tan x-x \right]}}$ $\left[ \tan \left( \frac{\pi }{4} \right)-\frac{\pi }{4}-\tan (0)-0 \right]=\left[ 1-\frac{\pi... read more$\int{_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\cos e{{c}^{2}}x-1)dx=[-\cot x-x]}=\left[ -\cot \left( \frac{\pi }{2} \right)-\frac{\pi }{2}+\cot \left( \frac{\pi }{4} \right)+\frac{\pi }{4} \right]$... read more$=[\tan \left( \frac{\pi }{6} \right)-\tan 0]=\frac{1}{\sqrt{3}}$Q.14$\int{_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\cos e{{c}^{2}}xdx=[-\cot x]}\left[ -\cot \left( \frac{\pi }{4} \right)+\cot...

$=[{{\sin }^{-1}}1-{{\sin }^{-1}}0]$ $=\frac{\pi }{2}$

$=[{{\tan }^{-1}}\infty {{\tan }^{-1}}0]$ $=\pi /2$

$=[{{\tan }^{-1}}1-{{\tan }^{-1}}0]$ $=\pi /4$

$=\left[ 3{{(8)}^{\frac{1}{3}}}-3{{(1)}^{\frac{1}{3}}} \right]$ $=\left[ 6-3 \right]$ $=3$

$=\left[ \frac{3}{2}{{1}^{\frac{4}{3}}}-0 \right]$ $=\frac{3}{2}$

$=\left[ 2\sqrt{4-2} \right]$ $=\left[ 4-2 \right]$ $=2$

$=\left[ \log (-1)-\log (-4) \right]$ $=-\left[ \log (-4)-\log (-1) \right]$ $-\left[ \log \left( \frac{-4}{-1} \right) \right]$ $=-\log 4$

$=\frac{4}{7}\left[ {{16}^{\frac{7}{4}}}-1 \right]$ $=\frac{512}{7}$

$\frac{{{2}^{-4}}}{-4}-\frac{1}{-4}$ $\frac{16-1}{64}$ $\frac{15}{64}$
$\frac{2}{3}\left[ {{4}^{\frac{3}{2}}}-1 \right]$ $=\frac{14}{3}$
$=\frac{{{3}^{5}}}{5}-\frac{1}{5}$ $\frac{243-1}{5}$ $=\frac{243}{5}$