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$$\int x\sec^2(x)\tan(x)\,dx$$ I just want to know what trigonometric function I need to use. I'm trying to integrate by parts. My book says that the integral equals $${x\over2\cos^2(x)}-{\sin(x)\over2\cos(x)}+C$$

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    $\begingroup$ I'm not quite sure what you're asking for when you ask about "what trigonometric function I need to use". Could you clarify what you mean there? $\endgroup$ – Omnomnomnom Oct 25 '18 at 17:31
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$$I=\int x\sec^2(x)\tan(x)dx$$ I will be walking you through this integral step-by-step.

First, we integrate by parts: $$u=x\\du=dx$$ And $$dv=\sec^2(x)\tan(x)dx$$ Which is an integral I'd like to demonstrate $$v=\int dv\\v=\int\sec^2(x)\tan(x)dx$$ For this, we will use a $\omega$-substitution. Let $\omega=\tan(x)$. Therefore, $d\omega=\sec^2(x)dx$, giving $$v=\int\omega d\omega=\frac{\omega^2}{2}\\v=\frac{\tan^2(x)}{2}$$ Next we finish our integration by parts: $$I=uv-\int vdu\\I=\frac{x\tan^2(x)}{2}-\int\frac{\tan^2(x)}{2}dx$$ Then we define $A=\int\frac{\tan^2(x)}{2}dx$, which gives $I=\frac{x\tan^2(x)}{2}-A$.

Here we go with the next integral: $$A=\int\frac{\tan^2(x)}{2}dx=\frac1{2}\int\tan^2(x)dx$$ Here we use a trig identity $\tan^2(x)=\sec^2(x)-1$. $$A=\frac1{2}\int(\sec^2(x)-1)dx=\frac1{2}\int\sec^2(x)dx-\frac1{2}\int dx$$ $$A=\frac{\tan(x)-x}{2}$$ Thus, $$I=\frac{x\tan^2(x)-(\tan(x)-x)}2$$ $$I=\frac{x\tan^2(x)-\tan(x)+x}2$$ And like always, add your constant: $$I=\frac{x\tan^2(x)-\tan(x)+x}2+C$$

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First you are going to need to use integration by parts... this will get the "x" out of the integral.

$\int x\sec^2 x\tan x \ dx$

$u = x, dv = \sec^2 x\tan x \ dx\\ du = dx, v = \frac 12 \sec^2 x$

How did I get v?

$v = \int \sec^2 x\tan x\ dx$

Now you will need to do a substitution $w = \tan x, dw = \sec^2 x$ or $w = \sec x, dw = \sec x\tan x$ either will work.

$\int x\sec^2 x\tan x \ dx\\ \frac 12 x\sec^2 x - \frac 12 \int \sec^2 x\ dx\\ \frac 12 x\sec^2 x - \frac 12 \tan x + c$

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Given $$\int x\sec^2(x)\tan(x)\ dx$$

Apply Integration By Parts $u=x$ and $v^{\prime}=\sec^2(x)\tan(x)$

and you get $$\dfrac12\tan^2(x)-\int\dfrac12\tan^2(x)\ dx$$

Can you take it from here?

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  • $\begingroup$ Thanks for your reply 😀 $\endgroup$ – Henry Trancer Oct 25 '18 at 22:10
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Let $u = x, dv = \sec^2 x\tan xdx\implies v = \displaystyle \int\sec^2 x\tan xdx= \displaystyle \int \tan xd(\tan x)= \displaystyle \int wdw, w = \tan x$. Can you put it together ?

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Hint: With $u = x$ and $dv = \sec^2 \tan(x)\,dx$, we have $$ \int u \,dv = uv - \int v \,du = \frac 12 x \tan^2 x - \int \tan^2 x\,dx $$ We can compute the integral of $\sec^2 \tan(x)$ with the $u$-substitution $u = \tan(x)$.

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