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I am stuck on this one:

$$ \int_0^{\frac{\pi}{4}} \frac{1 - \cos^2{\theta}}{\cos^{2}{\theta}} d \theta$$

$$ = \int_0^{\frac{\pi}{4}} \tan^{2} \theta$$

What is the antiderivative of $\tan^2{\theta}$

Is it this:

Since:

$$\frac{d}{dt} \tan{t} = \sec^2{t} = 1 + \tan{t}$$

then: $$ \left [ -t + \tan{t} \right ]_0^{\frac{\pi}{4}} $$

$$= \frac{-\pi}{4} + 1$$

But Wolfram alpha gets this:

enter image description here

Where did I go wrong?

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$$ \int_0^{\frac{\pi}{4}} \frac{1 - \cos^2{\theta}}{\cos^{2}{\theta}} d \theta = \int_0^{\frac{\pi}{4}} sec^2\theta d\theta - \int_0^ {\frac{\pi}{4}} 1 d\theta = \left[ tan\theta -\theta \right]_0^ {\frac{\pi}{4}} $$

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$$\frac{d}{dt}\tan t=\sec^2t=1+\tan^2 t$$ so an antiderivative of $\tan^2t$ is $-t+\tan t$.

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  • $\begingroup$ I see, and I forget, to find the integral, we only need one specific antiderivative of f right? $\endgroup$ – Jwan622 Apr 13 '18 at 6:24
  • $\begingroup$ any idea why my answer is different from WA? $\endgroup$ – Jwan622 Apr 13 '18 at 6:28
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    $\begingroup$ Your question has a minus sign where that entered into Wolfie has a plus sign. $\endgroup$ – Lord Shark the Unknown Apr 13 '18 at 6:30

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