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$\int \frac{\tan x}{\cos^2x}dx=\int\frac{\frac{\sin x}{\cos x}}{\cos^2 x}dx=\int \frac{\sin x}{\cos^3x}dx$
$\cos x=t, -\sin xdx=dt \Rightarrow \sin xdx=-dt$
$\int \frac{-dt}{t^3}=-\int t^{-3}dt=\frac{1}{2}t^{-2}dt=\frac{1}{2\cos^2x}+C $
$\int \frac{\tan x}{\cos^2x}dx\\\tan x=t,\frac{dx}{\cos^2 x}=dt\\\int t dt=\frac{1}{2}t^2+C=\frac{1}{2}\tan^2x+C$
which solution is correct?

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  • $\begingroup$ Here's another one to try. Do $\int x\sqrt{x+1}\,dx$ (a) by parts and (b) by substituting $u=\sqrt{x+1}$. Those answers don't look remotely "the same," do they? $\endgroup$ Dec 15, 2019 at 23:33
  • $\begingroup$ $\int x\sqrt{x+1}dx,u=\sqrt{x+1},u^2=x+1,x=u^2-1,\int(u^2-1)\sqrt{u^2-1+1}du,\int (u^3-u) du=\frac{1}{4}\left(\sqrt{x+1}\right)^4-\frac{1}{2}(x+1)+C,\int x\sqrt{x+1}dx=\int x\left(\frac{2}{3}(x+1)^{\frac{3}{2}}\right)'dx=x\left(\frac{2}{3}(x+1)^{\frac{3}{2}}\right)-\int \left(\frac{2}{3}(x+1)^{\frac{3}{2}}\right)=x\left(\frac{2}{3}(x+1)^{\frac{3}{2}}\right)-\frac{4}{15}(x+1)^\frac{5}{2}+C$ Am I counting this right? $\endgroup$
    – vmahth1
    Dec 15, 2019 at 23:58
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    $\begingroup$ You forgot $dx=2u\,du$ in the first one, right? Yours look a little too much different to possibly differ by a constant. :) $\endgroup$ Dec 15, 2019 at 23:59

1 Answer 1

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Both are correct, recall $\tan^{2}(x)+1=\sec^{2}(x)$, so that your two answers differ by a constant.

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