Question

Evaluate the integral

${\int }_0^{\pi }x{\sin }^{5}x{\cos }^{6}xdx=?$

• A. $\frac{5\pi }{16}$
• B. $\frac{35\pi }{128}$
• C. $\frac{5\pi }{8}$
• D. $\frac{8\pi }{693}$

Answer 1

The correct option is D $\frac{8\pi }{693}$

Let $I={\int }_0^{\pi }x{\sin }^{5}x\cdot {\cos }^{6}xdx$

$I={\int }_0^{\pi }(\pi –x)\cdot {\sin }^5(\pi –x)\cdot {\cos }^6(\pi –x)dx(\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b–x\left)dx\right)$

$I={\int }_0^{\pi }(\pi –x){\sin }^{5}x\cdot {\cos }^{6}xdx$

$I={\int }_0^{\pi }\pi {\sin }^{5}x\cdot {\cos }^{6}xdx-{\int }_0^{\pi }x\cdot {\sin }^{5}x\cdot {\cos }^{6}xdx$

$\therefore I=\pi {\int }_0^{\pi }{\sin }^{5}x\cdot {\cos }^{6}xdx–I$

$\therefore I=\frac{\pi }{2}{\int }_0^{\pi }{\sin }^{5}x\cdot {\cos }^{6}dx$

$I=2\cdot \frac{\pi }{2}{\int }_0^{\pi }{\sin }^{5}x\cdot {\cos }^{6}x\cdot dx$ $(\because f(2a–x)=f(x)\therefore {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx)$

$I=\pi \cdot \frac{4\times 2\times 5\times 3\times 1}{11\times 9\times 7\times 5\times 3\times 1}=\frac{8\pi }{693}$ (using walli’s formula).