Question

If ${\int }_0^{2\pi }x.\varphi (si{n}^{4}x+ta{n}^{2}x)dx=k{\int }_0^{\pi /2}\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

then the value of $k$ is?

• A. $\pi$
• B. $2\pi$
• C. $3\pi$
• D. $4\pi$

Answer 1

The correct option is D $4\pi$

${\int }_0^{2\pi }x\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$I={\int }_0^{2\pi }(2\pi -x)\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$={\int }_0^{2\pi }2\pi \varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$2I=2\pi {\int }_0^{2\pi }\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$2I=2\pi \left(4\right){\int }_0^{\frac{\pi }{2}}\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$I=4\pi {\int }_0^{\frac{\pi }{2}}\varphi (si{n}^{4}x+ta{n}^{2}x)dx$

$k=4\pi$