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प्रश्न
` ∫ tan x sec^4 x dx `
बेरीज
उत्तर
` ∫ tan x sec^4 x dx `
= ∫ tan x. sec2 x . sec^2 x dx
= ∫ tan x (1 + tan^2 x) sec^2 x dx
Let tan x = t
⇒ sec2 x dx = dt
Now, ∫ tan x (1 + tan2 x) sec2 x dx
= ∫ t (1 + t2) dt
= ∫ (t + t3) dt
\[= \frac{t^2}{2} + \frac{t^4}{4} + C\]
\[ = \frac{1}{2} \tan^2 x + \frac{1}{4} \text{ tan }^4 x + C \]
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