Question

integration sec^2x.cosec^4x

Answer 1

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Answer 2

Answer:

The integration of $Sec^2{x}.Cosec^4{x}$ is $-1/3Cot^3{x}+tan {x} +2cotx +c$.

Step-by-step explanation:

We have expression as $Sec^2{x}.Cosec^4{x}$ and we need to integrate it with respect to x which can be done as follows

$\int{sec^2{x}.Cosec^4x} dx \\\int\frac{sec^2x}{sin^4x}.dx$

Let us multiply $Cos^4x$ in numerator and denominator then we get

$\int\frac{sec^4.sec^4x}{tan^4x}.dx$

$\int\frac{sec^2x.(1-tan^2x)^2}{tan^4x}.dx$     {∵$sec^2x-tan^2x=1$}

Now let us say

$tan x = v$

⇒$sec^2x.dx=dv$

Then equation becomes

$\int\frac{(1-v^2)^2}{v^4}.dv\\\int\frac{1+v^4-2v^2}{v^4}.dv$

$\int\frac{1}{v^4}.dv+\int{1.dv}-\int\frac{2}{v^2}.dv\\\frac{-v^{-3}}{3}+v+2v^{-1}$

Now by putting the original value of v which is tan x we get

$\frac{-1}{3tan^3x}+tanx+\frac{2}{tanx}\\\frac{-1}{3}.Cotx+tan x+2Cotx+C$

Hence the integration $Sec^2{x}.Cosec^4{x}$ of  is $\frac{-1}{3}.Cotx+tan x+2Cotx+C$