Quick Antiderivative of Tan^2x and Sec^2x

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Homework Statement


Find the antiderivative of y=tan^2x+sec^2x


Homework Equations


N/A



The Attempt at a Solution


Seems to be a simple question, but the answer is eluding me no matter what I do. My first try was to replace the tan^2 with sec^2-1, and then factor out a 1/2 from the resulting 2sec^2x-1, but after that I have no idea how to continue. I know that the answer I should wind up with is Y=2tanx-x+C, but have no idea how to go about getting this answer :-p Any help would be greatly appreciated, thanks.
 
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Hint: d/dx(tan(x)) = sec2(x)
Every derivative formula gives you an antiderivative formula for free, if look at it the right way.

I also believe the answer you should end up with is tan(2x) - x + C, which is slightly different from what you show.
 
Ohh I see it now, thanks Mark, and no the answer is correct, as it is what I arrived at as well just now.
It might have been my mistake writing the question wrong, as it is y=(tanx)^2+(secx)^2. I just wrote it as y=tan^2x+sec^2x assuming people would know that, for example, tan^2x=(tanx)^2. I apologize for that mistake.
 
Yes, I misread what you wrote.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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