Question

${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\sin 2x}dx=$

• A. $\frac{1}{20}\log 5$
• B. $\frac{1}{20}\log 3$
• C. $\log 3$

Answer 1

The correct option is B $\frac{1}{20}\log 3$

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\sin 2x}dx$

Let $\sin x-\cos x=t$
Then, $(\sin x+\cos x)dx=dt$

When $x=0$, $t=\sin 0-\cos 0=-1$

When $x=\frac{\pi }{4}$, $t=\sin \frac{\pi }{4}-\cos \frac{\pi }{4}=0$

Then,
$I={\int }_{-1}^0\frac{dt}{9+16(1-t^2)}={\int }_{-1}^0\frac{dt}{25-16t^2}$
$=\frac{1}{10}{\int }_{-1}^0(\frac{1}{5-4t}+\frac{1}{5+4t})dt$
$={|\frac{1}{10}.\frac{1}{4}[\log (5+4t)-\log (5-4t)\left]\right|}_{-1}^0$
$=\frac{1}{40}(\log 9-\log 1)=\frac{1}{20}\log 3$