Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

The value of the integral $\int\limits_{0}^{π/4}\frac{\sin x +\cos x}{3+ \sin 2x}dx$, is

Options:

$\log 2$

$\log 3$

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

$\frac{1}{8}\log 3$

Correct Answer:

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

Explanation:

We have,

$I=\int\limits_{0}^{π/4}\frac{\sin x +\cos x}{3+ \sin 2x}dx=-\int\limits_{0}^{π/4}\frac{\sin x +\cos x}{(\sin x -\cos x)^2-4}dx$

$⇒I=\int\limits_{0}^{π/4}\frac{\sin x +\cos x}{3+ \sin 2x}dx=-\int\limits_{-1}^{0}\frac{dt}{t^2-2^2}$, where $t = \sin x-\cos x$

$⇒I=\int\limits_{0}^{π/4}\frac{\sin x +\cos x}{3+ \sin 2x}dx=-\frac{1}{4}\left[\log\left|\frac{t-2}{t+2}\right|\right]_{-1}^{0}=\frac{1}{4}\log 3$