How do you find the integral of #(1+ tan^2x)sec^2xdx#?

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Answer 1 · 2015-04-13T17:37:42

#int(1+tan^2(x))sec^2(x)dx#

We know :

#1+tan^2(x) = 1/cos^2(x)#

#sec^2(x) = 1/cos^2(x)#

So we have : #int1/cos^4(x)dx#

Let's #t = tan(x) and dt = 1/cos^2(x)dx#

We have :

#int1/cos^2(x)dt#

#int1+tan^2dt#

#int1+t^2dt#

#[t+1/3t^3]#

#[tan(x)+1/3tan^3(x)]+C#