Question

Evaluate: ${\int }_0^{\pi }{\sin }^{6}xdx$

• A. $\frac{5\pi }{16}$
• B. $\frac{35\pi }{128}$
• C. $\frac{5\pi }{8}$
• D. $\frac{5\pi }{18}$

Answer 1

The correct option is A $\frac{5\pi }{16}$

$I={\int }_0^{\pi }{\sin }^{6}xdx$

Now, If $f(2a-x)=f\left(x\right)$, then ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

$⟹I=2{\int }_0^{\pi /2}{\sin }^{6}xdx=I_1\left(say\right)$

$I=2{\int }_0^{\pi /2}{\sin }^6(\frac{\pi }{2}-x)dx$

$=2{\int }_0^{\pi /2}{\cos }^{6}xdx=I_2$

Hence, $2I=2{\int }_0^{\pi /2}({\sin }^{6}x+{\cos }^{6}x)dx$

$⟹I={\int }_0^{\pi /2}\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)]dx$

$⟹I={\int }_0^{\pi /2}(1-3{\sin }^{2}x{\cos }^{2}x)dx$

$⟹I={\int }_0^{\pi /2}(1-\frac{3}{4}(4{\sin }^{2}x{\cos }^{2}x\left)\right)dx$

$⟹I={\int }_0^{\pi /2}(1-\frac{3}{4}{\sin }^{2}2x)dx$

$⟹I={\int }_0^{\pi /2}(1-\frac{3}{4}.\frac{1-\cos 4x}{2})dx$

$⟹I={\int }_0^{\pi /2}(\frac{5}{8}+\frac{3}{8}\cos 4x)dx$

$⟹I=\frac{5}{8}{\left[x\right]}_0^{\pi /2}+\frac{3}{32}{\left[\sin 4x\right]}_0^{\pi /2}$

$⟹I=\frac{5\pi }{16}$

Hence, answer is option-(A).