Question

Evaluate the integral

${\int }_0^{\pi /4}\frac{sin\theta +cos\theta }{9+16sin2\theta }d\theta$

• A. $\frac{1}{20}\log 2$
• B. $\frac{1}{20}\log 3$
• C. $\log 3$
• D. $\log 2$

Answer 1

The correct option is B $\frac{1}{20}\log 3$

${\int }_0^{\pi /4}\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }d\theta$

Let $\sin \theta -\cos \theta =t$

$\therefore (\cos \theta +\sin \theta )d\theta =dt$

$={\int }_{-1}^0\frac{dt}{9+16(1–t^2)}(\because t^2=1–2\cdot \sin \theta \cdot \cos \theta )$

$={\int }_{-1}^0\frac{dt}{25–16t^2}$

$=\frac{1}{16}{\int }_{-1}^0\frac{dt}{{\left(\frac{5}{4}\right)}^2–t^2}$

$=\frac{1}{16}\times \frac{1}{2\times \frac{5}{4}}{\left[ln\left|\frac{\frac{5}{4}+t}{\frac{5}{4}–t}\right|\right]}_{-1}^0$

$(\because \int \frac{1}{a^2–x^2}dx=\frac{1}{2a}ln\left|\frac{a+x}{a–x}\right|+c)$

$=\frac{1}{40}[ln|1|-ln\left|\frac{\frac{1}{4}}{\frac{9}{4}}\right|]$

$=\frac{1}{40}ln9=\frac{ln3}{20}$

$\therefore {\int }_0^{\pi /4}\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }d\theta =\frac{ln3}{20}$.