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प्रश्न
` ∫ tan^5 x sec ^4 x dx `
बेरीज
उत्तर
` ∫ 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
\[= \frac{t^6}{6} + \frac{t^8}{8} + C\]
\[ = \frac{\tan^6 x}{6} + \frac{\tan^8 x}{8} + C\]
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