Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{7+9\sin 2x}dx$

• A. $-\frac{\log 3}{4}$
• B. $-\frac{\log 3}{36}$
• C. $-\frac{\log 7}{12}$
• D. $-\frac{\log 7}{24}$

Answer 1

The correct option is D $-\frac{\log 7}{24}$

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{7+9\sin 2x}dx$

Let $\sin x-\cos x=u\Rightarrow (\cos x+\sin x)dx=du$

Also $(\sin x-\cos x{)}^2=u^2$

$\Rightarrow 1-\sin 2x=u^2\Rightarrow \sin 2x=1-u^2$

And

upper limit $=0$, lower limit $=-1$

Ergo

$I={\int }_{-1}^0\frac{du}{7+9(1-u^2)}={\int }_0^1\frac{du}{9u^2-16}$

$=\frac{1}{9}{\int }_0^1\frac{du}{u^2-{\left(\frac{4}{3}\right)}^2}=\frac{1}{9}\cdot \frac{1}{2\cdot \frac{4}{3}}\log \left|\frac{x-\frac{4}{3}}{x+\frac{4}{3}}\right|{]}_0^1=-\frac{\log 7}{24}$