Q111 / 180 · KCET 2010 · Mathematics
$\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is
- Correct answerA$\frac{1}{4} \log 3$
- B$\log 3$
- C$\frac{1}{2 \log 3}$
- D$2 \log 3$
Solution
$\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
$=\int_{0}^{\pi / 4} \frac{(\sin x+\cos x)}{4-(1-\sin 2 x)} d x$
$=\int_{0}^{\pi / 4} \frac{(\sin x+\cos x)}{4-(\sin x-\cos x)^{2}} d x$
Put $\quad t=(\sin x-\cos x)$
$d t=(\cos x+\sin x) d x$
$=\int_{-1}^{0} \frac{d t}{\left(4-t^{2}\right)}$
$=\int_{-1}^{0} \frac{\mathrm{dt}}{(2+\mathrm{t})(2-\mathrm{t})}$
$=\frac{1}{4} \int_{-1}^{0}\left[\frac{1}{2+t}+\frac{1}{2-t}\right] d t$
$=\frac{1}{4}[\log (2+t)-\log (2-t)]_{-1}^{0}$
$$
\begin{aligned}
&=\frac{1}{4}\left[\log \left(\frac{2+\mathrm{t}}{2-\mathrm{t}}\right)\right]_{-1}^{0} \\
&=\frac{1}{4}\left[\log (1)-\log \left(\frac{1}{3}\right)\right] \\
&=\frac{1}{4} \log 3
\end{aligned}
$$
Hence, the correct answer is option: 'A'.