Question

Find $\stackrel{\pi /2}{\underset{0}{\int }}si{n}^{7}xdx$

Answer 1

Let $I=\int {\sin }^{7}xdx$

$\Rightarrow I=\int \sin x\cdot {\sin }^{6}xdx$

Integrating by parts, we have

$I=-\cos x\left({\sin }^{6}x\right)-\int (-\cos x)6{\sin }^{5}x\cos xdx$

$I=-{\sin }^{6}xcosx+6\int {\cos }^{2}x{\sin }^{5}xdx$

$I=-{\sin }^{6}x\cos x+6\int {\sin }^{5}x(1-{\sin }^{2}x)dx$

$I=-{\sin }^{6}x\cos x+6\int {\sin }^{5}xdx-6\int {\sin }^{7}xdx$

$I+6I=-{\sin }^{6}x\cos x+6\int {\sin }^{5}xdx$

$7I=-{\sin }^{6}x\cos x+6\int {\sin }^{5}xdx.....\left(1\right)$

Now, let

$I^{\prime }=\int {\sin }^{5}xdx$

$\Rightarrow I^{\prime }=\int \sin x\cdot {\sin }^{4}xdx$

Integrating by parts, we have

$I^{\prime }=-\cos x\left({\sin }^{4}x\right)-\int (-\cos x)4{\sin }^{3}x\cos xdx$

$I^{\prime }=-{\sin }^{4}xcosx+4\int {\cos }^{2}x{\sin }^{3}xdx$

$I^{\prime }=-{\sin }^{4}x\cos x+4\int {\sin }^{3}x(1-{\sin }^{2}x)dx$

$I^{\prime }=-{\sin }^{6}x\cos x+4\int {\sin }^{3}xdx-4\int {\sin }^{5}xdx$

$I^{\prime }+4I^{\prime }=-{\sin }^{6}x\cos x+4\int {\sin }^{3}xdx$

$5I^{\prime }=-{\sin }^{6}x\cos x+4\int {\sin }^{3}xdx$

$\Rightarrow I^{\prime }=-\frac{1}{5}{\sin }^{6}x\cos x+\frac{4}{5}\int {\sin }^{3}xdx$

Now, again let

$I^{\prime \prime }=\int {\sin }^{3}xdx$

$\Rightarrow I^{\prime \prime }=\int \sin x(1-{\cos }^{2}x)dx$

$I^{\prime \prime }=\int \sin x-\sin x{\cos }^{2}xdx$

$\Rightarrow I^{\prime \prime }=\int \sin xdx-\int \sin x{\cos }^{2}xdx$

$\Rightarrow I^{\prime \prime }=-\cos x+\frac{{\cos }^{3}x}{3}$

Now substituting the value of $I^{\prime }$ and $I^{\prime \prime }$ in equation $\left(1\right)$, we get

$7I=-{\sin }^{6}x\cos x+6[-\frac{1}{5}{\sin }^{6}x\cos x+\frac{4}{5}(-\cos x+\frac{{\cos }^{3}x}{3})]$

$I=-\frac{1}{7}{\sin }^{6}x\cos x-\frac{6}{35}{\sin }^{6}x\cos x-\frac{24}{35}\cos x+\frac{24}{105}{\cos }^{3}x$

Therefore,

${\int }_0^{\frac{\pi }{2}}{\sin }^{7}xdx={[-\frac{1}{7}{\sin }^{6}x\cos x-\frac{6}{35}{\sin }^{6}x\cos x-\frac{24}{35}\cos x+\frac{24}{105}{\cos }^{3}x]}_0^{\frac{\pi }{2}}$

$=(-\frac{1}{7}{\sin }^6\frac{\pi }{2}\cos x-\frac{6}{35}{\sin }^6\frac{\pi }{2}\cos \frac{\pi }{2}-\frac{24}{35}\cos \frac{\pi }{2}+\frac{24}{105}{\cos }^3\frac{\pi }{2})$

$-(-\frac{1}{7}{\sin }^{6}0\cos 0-\frac{6}{35}{\sin }^{6}0\cos 0-\frac{24}{35}\cos 0+\frac{24}{105}{\cos }^{3}0)$

$=\frac{24-72}{105}$

$=-\frac{48}{105}=-\frac{16}{35}$

Thus ${\int }_0^{\frac{\pi }{2}}{\sin }^{7}xdx=-\frac{16}{35}$