Question

Evaluate ${\int }_0^{\pi /2}\frac{\sin x+\cos x}{3+\sin 2x}dx$

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

Answer 1

The correct option is D $\frac{1}{4}{\log }_{e}3$

Let $I={\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+\sin 2x}dx$

$I={\int }_0^{\pi /4}\frac{\sin x+\cos x}{3+1-1+\sin 2x}dx$

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

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

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

Let $(\sin x-\cos x)=t\Rightarrow (\cos x+\sin x)=\frac{dt}{dx}$

$\therefore I_t=\int \frac{1}{4-(t{)}^2}dt$

$\therefore I=\frac{1}{4}{\left[\log \left|\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right|\right]}_0^{\pi /4}$

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

$=\frac{1}{4}{\log }_{e}3$