∫ Tan 3 X Sec 4 X D X - Mathematics

Sum
$\int \tan^3 x\ \sec^4 x\ dx$

Solution

$\text{ Let I }= \int \tan^3 x \cdot \sec^4 x\ dx$
$= \int \tan^3 x \cdot \sec^2 x \cdot \sec^2 x\ dx$
$= \int \tan^3 x \left( 1 + \tan^2 x \right) \cdot \sec^2 x\ dx$
$= \int\left( \tan^3 x + \tan^5 x \right) \sec^2 x\ dx$
$\text{Putting} \tan x = t$
$\Rightarrow \sec^2 x \text{ dx } = dt$
$\therefore I = \int \left( t^3 + t^5 \right) dt$
$= \frac{t^4}{4} + \frac{t^6}{6} + C$
$= \frac{\tan^4 x}{4} + \frac{\tan^6 x}{6} + C........... \left[ \because t = \tan x \right]$

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 82 | Page 204