First slide

Methods of integration

Question

$I_{n} = \int \tan^{n} x d x then I_{n} + I_{n + 2} is equal to$

Moderate

A

$\frac{\tan^{n - 1} x}{n - 1} + C$
B

$\frac{\tan^{n + 1} x}{n + 1} + C$
C

$\frac{\tan^{n} x}{n} + C$
D

$\frac{\tan^{2 n + 2}}{2 n + 2} + C$

Solution

$I_{n} + I_{n + 2} = \int \tan^{n} x d x + \int \tan^{n + 2} x d x$
$\begin{array}{l} = \int \tan^{n} x (\tan^{2} x + 1) d x \\ = \int \tan^{n} x \sec^{2} x d x \end{array}$
$Put \tan x = t$
$\begin{array}{l} \sec^{2} x d x = d t \\ = \int t^{n} d t \\ = \frac{t^{n + 1}}{n + 1} \\ = \frac{\tan^{n + 1}}{n + 1} + C \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App