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∫ Tan 5 X Sec 4 X D X - Mathematics

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Sum
` ∫  tan^5 x   sec ^4 x   dx `
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Solution

` ∫  tan^5 x   sec ^4 x   dx `

= ∫ tan5 x. sec2 x . sec2 x dx

= ∫ tan5 x (1 + tan2 x) sec2 x dx

Let tan x = t
⇒ sec2 x dx = dt

Now, ∫tan5x (1+tan2 x) sec2 x dx
= ∫ t5 (1 + t2) dt
= ∫ (t5 + t7) dt

\[= \frac{t^6}{6} + \frac{t^8}{8} + C\]
\[ = \frac{\tan^6 x}{6} + \frac{\tan^8 x}{8} + C\]

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.11 | Q 3 | Page 69

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