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∫ Tan 4 X D X - Mathematics

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Question

\[\int \tan^4 x\ dx\]
Sum
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Solution

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

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Chapter 19: Indefinite Integrals - Revision Excercise [Page 203]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Revision Excercise | Q 29 | Page 203

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