Question

If $\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{36+64\sin 2x}dx=\frac{\ln 3}{k},$ then the value of $k$ is

• A. $100$
• B. $80$
• C. $60$
• D. $40$

Answer 1

The correct option is D $40$

$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{36+64[1-(\sin x-\cos x{)}^2]}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x+\cos x}{100-64(\sin x-\cos x{)}^2}dx$
Put $8(\sin x-\cos x)=t8(\sin x+\cos x)dx=dt$
When $x=0\Rightarrow t=-8x=\frac{\pi }{2}\Rightarrow t=8$
$\Rightarrow I=\frac{1}{8}\underset{-8}{\overset{8}{\int }}\frac{dt}{100-t^2}$
$=\frac{2}{8}\underset{0}{\overset{8}{\int }}\frac{dt}{{10}^2-t^2}$ (even function)
$=\frac{1}{4}\times \frac{1}{2\times 10}\ln {\left|\frac{10+t}{10-t}\right|}_0^8=\frac{\ln 3}{40}\Rightarrow k=40$