First slide

Introduction to definite integration

Question

$\int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} dx$ is equal to

Easy

A

$\frac{π}{2}$
B

$\frac{π}{4}$
C

$\frac{3 π}{2}$
D

$π$

Solution

Let $I = \int_{0}^{π / 2} \frac{\sin x}{1 + \cos^{2} x} dx$
put $\cos x = t \Rightarrow - \sin xdx = dt \Rightarrow dx = - \frac{dt}{\sin x}$
For limit when $x = 0 \Rightarrow t = \cos 0 = 1 [∵ t = \cos x]$
and when $x = \frac{π}{2} \Rightarrow t = \cos \frac{π}{2} = 0$
$\begin{array}{l} ∴ I = \int_{1}^{0} \frac{\sin x}{1 + t^{2}} \cdot (\frac{dt}{- \sin x}) . \\ = - \int_{1}^{0} \frac{1}{1 + t^{2}} dt = \int_{0}^{1} \frac{1}{1 + t^{2}} dt \\ = {[\frac{1}{1} \tan^{- 1} (\frac{t}{1})]}_{1}^{0} [∵ \int \frac{1}{a^{2} + x^{2}} dx = \frac{1}{a} \tan^{- 1} \frac{x}{a}] \\ = - (0 - \frac{π}{4}) = \frac{π}{4} \end{array}$

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