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Question

\[\int \sec^4    \text{ x   tan x dx} \]
Sum
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Solution

\[\int \sec^4    \text{x . tan x dx}\]
\[ = \int \sec^2 x . \sec^2 \text{x  tan x dx}\]
\[ = \int \left( 1 + \tan^2 x \right) \sec^2 \text{x . tan x dx}\]
\[ = \int \left( \tan x + \tan^3 x \right) \sec^2 \text{x dx}\]
\[\text{Let tan x} = t\]
\[ \Rightarrow \sec^2 \text{x dx} = dt\]
\[Now, \int \left( \tan x + \tan^3 x \right) \sec^2\text{ x dx}\]
\[ = \int\left( t + t^3 \right) dt\]
\[ = \frac{t^2}{2} + \frac{t^4}{4} + C\]
\[ = \frac{\tan^2 x}{2} + \frac{\tan^4 x}{4} + C\]

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Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 58]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 44 | Page 58

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