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Q16  Find the integrals of the functions  \tan ^ 4x

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best_answer

the given question can be rearranged using trigonometric identities

tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1

Therefore, the integration of \tan^4x = \\=\int \sec^2x.\tan^2x\ dx-\int\sec^2x\ dx+\int dx\\ =(\int \sec^2x.\tan^2x\ dx)-\tan x+x+C\\...................(i)
Considering only \int \sec^2x.\tan^2x\ dx 
let \tan x =t\Rightarrow \sec^2x\ dx =dt
 

\int \sec^2x\tan^2x\ dx = \int t^2\ dt = t^3/3=\frac{\tan^3x}{3}

now the final solution is,

\int \tan^4x =\frac{\tan^3x}{3}-\tan x+x+C

Posted by

manish

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