Question

Evaluate ${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\{1-{(\sin x-\cos x)}^2\}}dx$

• A. $\frac{1}{20}\left(\log \sqrt{3}\right)$
• B. $\frac{1}{10}\left(\log 3\right)$
• C. $\frac{1}{20}\left(\log 3\right)$
• D. $\frac{1}{20}(\log 3-\log 2)$

Answer 1

The correct option is C $\frac{1}{20}\left(\log 3\right)$

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{9+16\{1-{(\sin x-\cos x)}^2\}}dx$
$\therefore I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{25-16{(\sin x-\cos x)}^2}dx$,
Substitute $4(\sin x-\cos x)=t⟹4(\cos x+\sin x)dx=dt$
$\frac{1}{4}{\int }_{-4}^0\frac{dt}{25-t^2}$
$I=\frac{1}{4}.\frac{1}{2\left(5\right)}{\left(\log \left|\frac{5-t}{5+t}\right|\right)}_{-4}^0$
$=\frac{1}{40}[\log \left|\frac{5-0}{5+0}\right|-\log \left|\frac{5+4}{5-4}\right|]$
$=\frac{1}{40}[\log 1-\log \left(\frac{1}{9}\right)]$, (where $\log 1=0$)
$=\frac{1}{40}\left[\log 9\right]=\frac{2}{40}\log 3=\frac{1}{20}\left(\log 3\right)$