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\[\Rightarrow ~1\text{ }=\text{ }\left( A\text{ }+\text{ }B \right)\text{ }x\text{ }+\text{ }\left( 3A\text{ }\text{ }2B \right)\] $⇒$ Then  $A\;+\;B\;=\;0\;… (1)$ And $3A\;–\;2B\;=\;1\;… (2)$...

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Solution: Given that $\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{3 / 2}} d x$ Putting $x=a \sin \theta$, then $d x=a \cos \theta d \theta$ and $\theta=\sin ^{-1}(x / a)$ The above equation becomes,...

Solution: As power of $\sin$ is odd, put $\cos x=t$ Therefore $\mathrm{dt}=-\sin \mathrm{x} \mathrm{d} \mathrm{x}$ On substitute these in above equation, we get $\begin{array}{l} \int \sin ^{3} x...

Solution: Assume $\sin x=t$ Therefore $\mathrm{d}(\sin \mathrm{x})=\mathrm{dt}=\cos \mathrm{x} \mathrm{dx}$ Putting $t=\sin x$ and $d t=\cos x d x$ in given equation, we get $\int \sin ^{5} x \cos x...

Solution: We can write the given question as $\begin{array}{l} \int \cos ^{5} x d x=\int \cos ^{3} x \cos ^{2} x d x \\ =\int^{\cos ^{3} x\left(1-\sin ^{2} x\right) d x}\left\{\text { since } \sin...

Solution: We can write the given equation as $\begin{array}{l} \int \sin ^{5} x d x=\int \sin ^{3} x \sin ^{2} x d x \\ =\int \sin ^{3} x\left(1-\cos ^{2} x\right) d x\left\{\text { since } \sin...

Solution: Suppose $\sin x=t$ It is known that the Differentiation of $\sin x=\cos x$ $\mathrm{dt}=\mathrm{d}(\sin \mathrm{x})=\cos \mathrm{x} \mathrm{dx}$ $\mathrm{Therefore},...

Solution: Assume $I=\int \sqrt{\tan x} \sec ^{4} x d x$ We can write the above equation as $\Rightarrow I=\int \sqrt{\tan x} \sec ^{2} x \sec ^{2} x d x$ Now, taking common $\begin{array}{l}...

Solution: Assume $I=\int \tan ^{5} x d x$ We can write the above equation as $\Rightarrow I=\int \tan ^{2} x \tan ^{3} x d x$ By using the standard formula $\Rightarrow I=\int\left(\sec ^{2}...

Solution: Assume $I=\int \sec ^{6} x \tan x d x$ We can write the above equation as $\Rightarrow I=\int \sec ^{5} x(\sec x \tan x) d x$ Substituting, $\sec x=t \Rightarrow \sec x \tan x d x=d t$, we...

Solution: Assume $I=\int \tan ^{5} x \sec ^{4} x d x$ We can write the above equation as $\Rightarrow I=\int \tan ^{5} x \sec ^{2} x \sec ^{2} x d x$ Taking $\tan ^{5} \mathrm{x}$ as common, we get...

Solution: Assume $I=\int \tan x \sec ^{4} x d x$ We can write the above equation as $\Rightarrow \mathrm{I}=\int \tan \mathrm{x} \sec ^{2} \mathrm{x} \sec ^{2} \mathrm{x} \mathrm{dx}$...

Solution: Assume $I=\int \tan ^{3} x \sec ^{2} x d x$ Assume $\tan \mathrm{x}=\mathrm{t}$, then $\Rightarrow \sec ^{2} x d x=d t$ On substituting the values of $x$, we get $\Rightarrow...

Solution: Assume $I=\int\left(2 x^{2}+3\right) \sqrt{x+2} d x$ Substituting $x+2=t \Rightarrow d x=d t$ On substituting the values of $x$ in given equation, we obtain $\begin{array}{l} \Rightarrow...

Solution: Assume $I=\int \frac{2 x-1}{(x-1)^{2}} d x$ Substituting $x-1=t \Rightarrow d x=d t$ On substituting the values of $x$, we get $\Rightarrow \mathrm{I}=\int...

Solution: Assume $\mathrm{I}=\int \frac{\mathrm{x}^{2}}{\sqrt{3 \mathrm{x}+4}} \mathrm{dx}$ Substituting $3 x+4=t \Rightarrow 3 d x=d t$ By substituting the values of $x$, we get $\Rightarrow I=\int...

Solution: Assume $I=\int \frac{x^{2}}{\sqrt{x-1}} d x$ By substituting $x-1=t \Rightarrow d x=d t$ On substituting the values we obtain $\Rightarrow \mathrm{I}=\int...

Solution: Assume $I=\int x^{2} \sqrt{x+2} d x$ By substituting, $x+2=t \Rightarrow d x=d t$ $\begin{array}{l} I=\int(t-2)^{2} \sqrt{t} d t \\ \Rightarrow I=\int\left(t^{2}-4 t+4\right) \sqrt{t} d t...

Solution: Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{d} \mathrm{t} \\ \Rightarrow...

Solution: Let $x-\cos x=t$ $\begin{array}{l} \Rightarrow d(x-\cos x)=d t \\ \Rightarrow(1+\sin x) d x=d t \end{array}$ $\therefore$ By substituting $\mathrm{t}$ and dt in given equation we obtain...

Solution: Let $\sin ^{-1} \mathrm{x}=\mathrm{t}$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(\sin ^{-1} \mathrm{x}\right)=\mathrm{dt} \\ \Rightarrow...

Solution: Let $\cot x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cot x)=d t \\ \Rightarrow-\operatorname{cosec}^{2} x \cdot d x=d t \\ \Rightarrow d x=\frac{-d t}{\csc ^{2} x} \end{array}$...

Solution: Let $1+\mathrm{e}^{\mathrm{x}}=\mathrm{t}$ $\Rightarrow d\left(1+e^{x}\right)=d t$ $\Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$ $\therefore$ By substituting $t$ and $dt$...

Solution: Let $\cos x=t$ $\begin{array}{l} \Rightarrow \mathrm{d}(\cos x)=d t \\ \Rightarrow-\sin x d x=d t \\ \Rightarrow d x=\frac{-d t}{\sin x} \end{array}$ $\therefore$ On substituting...

Solution: Let $1+e^{x}=t$ $\begin{array}{l} \Rightarrow \mathrm{d}\left(1+\mathrm{e}^{\mathrm{x}}\right)=\mathrm{dt} \\ \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}=\mathrm{d}...

Solution: Let $1+v x=t$ $\begin{array}{l} \Rightarrow d(1+v x)=d t \\ \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\ \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t \end{array}$ $\therefore$ On substituting...

Solution: Let $\log \left(1+\frac{1}{\mathrm{x}}\right)=\mathrm{t}$ $\begin{array}{l} \Rightarrow \operatorname{d}\left(\log \left(1+\frac{1}{\mathrm{x}}\right)\right)=\mathrm{dt} \\ \Rightarrow...

Solution: Let $\log x=t$ $\begin{array}{l} \Rightarrow d(\log x)=d t \\ \Rightarrow \frac{1}{x} d x=d t \end{array}$ By substituting $\mathrm{t}$ and $dt$ in above equation we obtain...

Solution: It is better to eliminate the denominator, in order to solve these equations. $\Rightarrow \int \frac{\sin (x-a)}{\sin (x-b)} d x$ Now, add and subtract $b$ in $(x-a)$ $\begin{array}{l}...

Solution: Suppose $\mathrm{I}=\int \frac{\cos 2 x}{(\cos \mathrm{x}+\sin \mathrm{x})^{2}} d x$ On substituting the formula, we obtain $=\int \frac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x$...

Solution: First of all we need to convert sec $x$ in terms of $\cos x$ It is known that $\Rightarrow \sec x=\frac{1}{\cos x}, \sec 2 x=\frac{1}{\cos 2 x}$ So, the above equation becomes,...

Solution: Given that, $\int \frac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} d x$ It is known that $\begin{array}{l} 1-\operatorname{Cos} x=2 \sin ^{2} \frac{x}{2} \\ 1+\cos x=2 \cos ^{2} \frac{x}{2}...

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Solution: Given that, $\int \frac{\sqrt{1+\cos 2 x}}{\sqrt{1-\cos 2 x}} d x$ It is known that $\begin{array}{l} 1+\cos 2 x=2 \cos ^{2} x \\ 1-\cos 2 x=2 \sin ^{2} x \end{array}$ On substituting...

Solution: Given that $\int \frac{1}{\sqrt{1-\cos 2 x}} d x$ In the equation given $\cos 2 x=\cos ^{2} x-\sin ^{2} x$ Also it is known that $\cos ^{2} x+\sin ^{2} x=1$ On substituting the values in...

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