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 :

$int1/cos^2(x)dt$

$int1+tan^2dt$

$int1+t^2dt$

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

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

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