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

Answer 1

Charles Anctil

#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#

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Answer 2

Isabella Ackerman

To integrate ( \int (\theta^2 + \sec^2\theta) , d\theta ), you can use the following steps:

Rewrite ( \sec^2\theta ) as ( 1 + \tan^2\theta ).
Distribute ( \theta^2 ) over ( 1 + \tan^2\theta ).
Integrate term by term.

The result will be:

[ \int (\theta^2 + \sec^2\theta) , d\theta = \frac{\theta^3}{3} + \tan\theta + C ]

where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.