Question

Question #61aa2

Answer 1

`inttan^3xsec^2xdx=1/4tan^4x+C`

Explanation:

`inttan^3xsec^2xdx`

the key here is to remember that

`d/(dx)(tan^nx)=ntan^(n-1)xsec^2x`

**see below**

so comparing this with the required integral

try `d/(dx)(tan^4x)`

`=4tan^3xsec^2x`

`inttan^3xsec^2xdx=1/4tan^4x+C`

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proof

`d/(dx)(tan^nx)=ntan^(n-1)xsec^2x`

`y=tan^nx`

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

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

by the chain rule

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

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

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

`=ntan^(n-1)xsec^2x`