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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 18

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Answer: $\frac{1}{-3\left ( 5+\tan x \right )^{3}}+c$

Hint: You must know about the integral rule of trigonometric functions.

Given: $\int \frac{\sec^{2}x}{\left ( 5+\tan x \right )^{4}}dx$

Solution: $\frac{\sec^{2}x}{\left ( 5+\tan x \right )^{4}}$

$t=5+\tan x$ differentiate both sides, $\left [ \frac{d}{dx}\tan x= \sec^{2} x \right ]$

$dt=\sec^{2}xdx$

$I=\int \frac{\sec^{2}x}{\left ( 5+\tan x \right )^{4}}dx$

$= \int \frac{dt}{t^{4}}$

$= \int t^{-4}dt$

$=\frac{t^{-3}}{-3}+c$

$=\frac{-1}{3t^{3}}+c$ $\left [ \int x^{n} dx= \frac{x^{n+1}}{n+1}\right ]$

$\therefore I=\frac{1}{-3\left ( 5+\tan x \right )^{3}}+c$

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