Question

What is the integral of `int tan^6(x)sec^4(x)dx`?

Answer 1

`=1/7tan^7x+1/9tan^9x+C`

Explanation:

for this question it is important to recognise that

`d/dx(tanx)=sec^2x`

and in general `color(blue)(d/dx(tan^nx)=ntan^(n-1)xsec^2x)`

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**(proof:
this is by use of the chain rule

`u=tanx=>(du)/(dx)=sec^2x`

`y=u^n=>(dy)/(du)=n u^(n-1)`

`(dy)/(dx)=(dy)/(du)xx(du)/(dx)`

`(dy)/(dx)=n u^(n-1)xxsec^2x=ntan^(n-1)xsec^2x`**

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so we re-write the integral into a form that uses this result.

`inttan^6xsec^4xdx`

`=inttan^6xsec^2x(1+tan^2x)`dx#

`=int(tan^6xsec^2x+tan^8xsec^2x)dx`

using the results in reverse.

`=1/7tan^7x+1/9tan^9x+C`