First slide

Introduction to integration

Question

$Evaluate \int \frac{\sqrt{\tan x}}{\sin x \cos x} d x$

Easy

A

$\sqrt{\tan x} + C$
B

$\frac{1}{2} \sqrt{\tan x} + C$
C

$2 \sqrt{\tan x} + C$
D

None of these

Solution

$\begin{array}{l} \int \frac{\sqrt{\tan x}}{\sin x \cos x} d x = \int \frac{\sqrt{\tan x} \frac{1}{\cos^{2} x}}{\frac{\sin x \cos x}{\cos^{2} x}} d x \\ = \int \frac{\sqrt{\tan x} \sec^{2} x}{\tan x} d x \\ = \int (\tan x)^{- 1 / 2} \sec^{2} x d x \\ = \frac{(\tan x)^{1 / 2}}{1 / 2} + C \\ = 2 \sqrt{\tan x} + C \end{array}$

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