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

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Answer 1 · 2016-12-05T21:26:26

#theta^3/3+tantheta+C#

#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:

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#