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Question
` ∫ tan^5 x dx `
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Solution
∫ tan5 x dx
= ∫ tan4 x. tan x dx
= ∫(sec2 x – 1)2 . tan x dx
= ∫ (sec4 x – 2 sec2 x + 1) tan x dx
= ∫ tan x . sec4 x dx – 2 ∫ sec2 x . tan x dx+ ∫ tan x dx
= ∫ sec2 x. sec2 x . tan x dx – 2 ∫ tan x sec2 x dx + ∫ tan x dx
= ∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx + ∫ tan x dx
Let I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
And I2=∫ tan x dx
∫ tan5 x dx=I1 + I2
Now, I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
Let tan x = t
⇒ sec2x dx = dt
I1=∫ (1 + tan2 x) . tan x . sec2 x dx – 2 ∫ tan x . sec2 x dx
∫ (1 + t2) . t. dt – 2 ∫ t. dt
∫ (t + t3) dt – 2 ∫ t dt
\[\frac{t^2}{2} + \frac{t^4}{4} - \frac{2 t^2}{2} + C_1 \]
\[ = \frac{t^4}{4} - \frac{t^2}{2} + C_1 \]
\[ = \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + C_1\]
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