Question

${\int }_0^{\pi /4}\frac{\sin x+\cos x}{7+9\sin 2x}dx=$

• A. $\frac{\log 3}{4}$
• B. $\frac{\log 3}{36}$
• C. $\frac{\log 3}{12}$
• D. $\frac{\log 7}{24}$

Answer 1

The correct option is D $\frac{\log 7}{24}$

${\int }_0^{\pi /4}\frac{\sin x+\cos x}{7+9\sin 2x}dx$

$\displaystyle \int _{ 0 }^{ \pi /4 }{ \cfrac { \sin { x } +\cos { x } }{ 16-9(\sin { x }-\cos{x})^{2} } } dx$

let $(\sin x-\cos x)=t$

$(\cos x+\sin x)dx=dt$

and limits are

$x=0\Rightarrow t=-1$

$x=\frac{\pi }{4}\Rightarrow t=0$

now,

$\displaystyle \int _{-1 }^{ 0 }{ \cfrac { 1 }{ 16 - 9t^{2} } } dt$

$\Rightarrow \frac{1}{24}{\left[\log \right(\frac{4+3(\sin x-\cos x)}{4-3(\sin x-\cos x)}\left)\right]}_0^{\frac{\pi }{4}}$

$\Rightarrow \frac{\log \left(7\right)}{24}$