Question

How do you integrate `int (theta^2+sec^2theta)d theta`?

Answer 1

`theta^3/3+tantheta+C`

Explanation:

`I=int(theta^2+sec^2theta)d theta`

Split up the integral:

`I=inttheta^2d theta+intsec^2thetad theta`

Think of the following two things:

• `intx^ndx=x^(n+1)/(n+1)+C`
• `d/dxtanx=sec^2x=>intsec^2xdx=tanx+C`

Don't be tripped up by the use of `theta` as a variable—this process is identical to integrating `int(x^2+sec^2x)dx`.

Thus:

`I=theta^3/3+tantheta+C`