Question

` ∫  tan^5 x   sec ^4 x   dx `

Answer 1

` ∫  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

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