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method 1:Substitution

$\int \sec x\tan x dx=\int \frac{\sin x}{\cos ^2x}dx$

Let $u=\cos x \implies -du=\sin x dx$

$-\int \frac{1}{u^2}du=\frac{1}{u}+c$

So $\int \sec x\tan x dx=\frac{1}{\cos x}=\sec x+c$

Method 2:integration by parts

$\int \sec x\tan x dx=\int \sec ^2x\sin xdx$

Let $u=\sin x$ and $dv=\sec^2x dx \implies du=\cos x dx$ and $v=\tan x$

$\int \sec ^2x\sin xdx=\tan x \sin x-\int \tan x\cos x dx$

$=\tan x \sin x-\int \sin x dx$

$=\tan x \sin x+\cos x +c$

I cannot tell where I went wrong especially with the second method.

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  • $\begingroup$ Always be aware that different methods can give different-looking expressions for the same result (or - for indefinite integration - the same result up to an additive constant, which is irrelevant because of the $+c$), especially if trigonometry gets involved because there are lots of trigonometric identities. $\endgroup$ – J.G. Apr 3 at 12:04
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Nothing went wrong. Just note that\begin{align}\tan(x)\sin(x)+\cos(x)&=\frac{\sin^2(x)}{\cos(x)}+\frac{\cos^2(x)}{\cos(x)}\\&=\frac1{\cos(x)}\\&=\sec(x).\end{align}

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