Question

Integral ${\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+\sin 2x}$ is equal to ?

• A. $\log 2$
• B. $\log 3$
• C. $-\frac{1}{4}\log \frac{1}{3}$
• D. $14\log 3$

Answer 1

Answer: (C)

$-\frac{1}{4}\log \frac{1}{3}$

$I={\int }_0^{\pi /4}\frac{\sin x+\cos x}{4-(1-\sin 2x)}dx$

$={\int }_0^{\pi /4}\frac{\sin x+\cos x}{4-(\sin x-\cos x{)}^2}dx$

$\sin x-\cos x=t$

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

When $x=0\Rightarrow \sin 0-\cos \theta =0-1=-1$

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

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

$=\frac{1}{2\times 2}{\left[ln\left|\frac{2+t}{2-t}\right|\right]}_{-1}^0$

$=\frac{1}{4}[ln|1|-ln\left|\frac{1}{3}\right|]$

$=\frac{-1}{4}ln\left[\frac{1}{3}\right]$