Question

$\stackrel{\pi /4}{\underset{0}{\int }}\frac{\sin x+\cos x}{9+16\sin 2x}dx$ is equal to

• A. $log3$
• B. $log2$
• C. $\frac{1}{40}log9$
• D. $\frac{1}{20}log3$

Answer 1

The correct option is C $\frac{1}{40}log9$

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

substitute $t=\sin x-\cos x\Rightarrow 1-t^2=\sin 2x\to dt=(\sin x+\cos x)dx$

$x(0,\frac{\pi }{4})\to t(-1,0)$

$\therefore {\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\sin 2x}dx={\int }_{-1}^0\frac{dt}{9+16(1-t^2)}$

$={\int }_{-1}^0\frac{dt}{9+16-16t^2}$

$={\int }_{-1}^0\frac{dt}{25-16t^2}$

$={\int }_{-1}^0\frac{dt}{(5{)}^2-(4t{)}^2}$

$=\frac{1}{4}\frac{1}{2\left(5\right)}{\left[\log \left(\frac{5+4t}{5-4t}\right)\right]}_{-1}^0$

$=\frac{1}{40}\left[\log \right(1)-\log \frac{1}{9}]$

$=\frac{1}{40}\log 9$