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},...