Question

Let $u={\int }_0^{\frac{\pi }{2}}\cos \left(\frac{2\pi }{3}{\sin }^{2}x\right)dx$ and $v={\int }_0^{\frac{\pi }{2}}\cos \left(\frac{\pi }{3}\sin x\right)dx$, then the relation between $u$ and $v$ is

• A. $2u=v$
• B. $2u=3v$
• C. $u=v$
• D. $u=2v$

Answer 1

Answer: (A)

$2u=v$

$u={\int }_0^{\frac{\pi }{2}}\cos \left(\frac{2\pi }{3}{\sin }^{2}x\right)dx$
$u={\int }_0^{\frac{\pi }{2}}\cos \left(\frac{2\pi }{3}{\cos }^{2}x\right)dx$
$\Rightarrow 2u={\int }_0^{\pi /2}(\cos \left(\frac{2\pi }{3}{\sin }^{2}x\right)+\cos \left(\frac{2\pi }{3}{\cos }^{2}x\right))dx$
$\Rightarrow 2u={\int }_0^{\pi /2}2\cos \frac{\pi }{3}cos(\frac{\pi }{3}\cos 2x)dx$
$\Rightarrow =\frac{1}{2}{\int }_0^{\pi /2}\cos \left(\frac{\pi }{3}\cos 2x\right)dx$
Put $2x=t$
$\Rightarrow 2dx=dt$
$\Rightarrow u=\frac{1}{4}{\int }_0^{\pi }\cos \left(\frac{\pi }{3}\cos t\right)dt$
$\Rightarrow u=\frac{1}{2}{\int }_0^{\pi /2}\cos \left(\frac{\pi }{3}\cos t\right)dt$
$\Rightarrow u=\frac{1}{2}{\int }_0^{\pi /2}\cos \left(\frac{\pi }{3}\sin t\right)dt$
$\Rightarrow 2u=v$