Question

The solution of the integral ${\int }_0^{2\pi }{\sin }^{4}x\cdot {\cos }^{6}xdx$, is

• A. $\frac{3\pi }{128}$
• B. $\frac{\pi }{128}$
• C. $\frac{5\pi }{128}$
• D. $\frac{7\pi }{128}$

Answer 1

The correct option is A $\frac{3\pi }{128}$

Let $I={\int }_0^{2\pi }{\sin }^{4}x\cdot {\cos }^{6}xdx$
$\Rightarrow I=2{\int }_0^{\pi }{\sin }^{4}x\cdot {\cos }^{6}xdx$ ($\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{2a}f\left(x\right)dx\text{if}f\left(x\right)iseven$)
$\Rightarrow I=4{\int }_0^{\pi /2}{\sin }^{4}x\cdot {\cos }^{6}xdx$ ($\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{2a}f\left(x\right)dx\text{if}f\left(x\right)iseven$)
$\Rightarrow I=4{\int }_0^{\pi /2}{\left(\frac{1-\cos 2x}{2}\right)}^2{\left(\frac{1+\cos 2x}{2}\right)}^{3}dx$
$\Rightarrow I=\frac{1}{8}{\int }_0^{\pi /2}{(1-\cos 2x)}^2{(1+\cos 2x)}^2(1+\cos 2x)dx$
$\Rightarrow I=\frac{1}{8}{\int }_0^{\pi /2}{\sin }^{4}2x(1+\cos 2x)dx$
$\Rightarrow I=\frac{1}{8}[{\int }_0^{\pi /2}{\sin }^{4}2xdx+{\int }_0^{\pi /2}{\sin }^{4}2x\cos 2xdx]$
Put $\sin 2x=t$ in second integral
$\Rightarrow 2\cos 2xdx=dt$
But , the limits of integral changes from 0 to 0 .
Hence, the value of second integral is 0
$\Rightarrow I=\frac{1}{8}{\int }_0^{\pi /2}{\left(\frac{1-\cos 4x}{2}\right)}^{2}dx$
$\Rightarrow I=\frac{1}{32}{\int }_0^{\pi /2}(1+{\cos }^{2}4x-2\cos 4x)dx$
$\Rightarrow I=\frac{1}{32}{\int }_0^{\pi /2}(1+\frac{1+\cos 8x}{2}-2\cos 4x)dx$
$\Rightarrow I=\frac{1}{32}{\int }_0^{\pi /2}(\frac{3}{2}+\frac{\cos 8x}{2}-2\cos 4x)dx$
$\Rightarrow I=\frac{1}{32}{[\frac{3x}{2}+\frac{\sin 8x}{16}-\frac{2\sin 4x}{4}]}_0^{\pi /2}$
$\Rightarrow I=\frac{3\pi }{128}$