`int sec^2(theta) tan^3(theta) d theta` Evaluate the indefinite integral.

You need to use the following substitution ` tan theta= t` , such that:

`tan theta= t=>(sec^2 theta)d theta= dt `

`int (sec^2 theta)tan^3 theta d theta = int t^3 dt`

`int t^3 dt = (t^4)/4 + c`

Replacing back `tan theta` for t yields:

`int (sec^2 theta)tan^3 theta d theta =(tan^4 theta)/4 + c`

Hence, evaluating the indefinite integral, yields `int (sec^2 theta)tan^3 theta d theta =(tan^4 theta)/4 + c`

You need to use the following substitution ` tan theta= t` , such that:

`tan theta= t=>(sec^2 theta)d theta= dt `

`int (sec^2 theta)tan^3 theta d theta = int t^3 dt`

`int t^3 dt = (t^4)/4 + c`

Replacing back `tan theta` for t yields:

`int (sec^2 theta)tan^3 theta d theta =(tan^4 theta)/4 + c`

Hence, evaluating the indefinite integral, yields `int (sec^2 theta)tan^3 theta d theta =(tan^4 theta)/4 + c`