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∫ Tan 5 X Sec 3 X D X - Mathematics

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Sum
\[\int \tan^5 x\ \sec^3 x\ dx\]
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Solution

\[\text{ Let I} = \int \tan^5 x \cdot \sec^3 x\ dx\]
\[ = \int \tan^4 x \cdot \sec^2 x \cdot \sec x \tan x\ dx\]
\[ = \int \left( \sec^2 x - 1 \right)^2 \cdot \sec^2 x \cdot \sec x \tan x\ dx\]
\[\text{ Putting  sec x} = t\]
\[ \Rightarrow \text{ sec x  tan x dx = dt}\]
\[ \therefore I = \int \left( t^2 - 1 \right)^2 \cdot t^2 \cdot dt\]
\[ = \int\left( t^4 - 2 t^2 + 1 \right) t^2 dt\]
\[ = \int\left( t^6 - 2 t^4 + t^2 \right) dt\]
\[ = \frac{t^7}{7} - \frac{2 t^5}{5} + \frac{t^3}{3} + C\]
\[ = \frac{1}{7} \sec^7 x - \frac{2}{5} \sec^5 x + \frac{1}{3} \sec^3 x + C................. \left[ \because t = \sec x \right]\]

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 81 | Page 204

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