# Integrate $\int x\sec^2(x)\tan(x)\,dx$

$$\int x\sec^2(x)\tan(x)\,dx$$ 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 $${x\over2\cos^2(x)}-{\sin(x)\over2\cos(x)}+C$$

• 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? – Omnomnomnom Oct 25 '18 at 17:31

$$I=\int x\sec^2(x)\tan(x)dx$$ 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: $$I=\frac{x\tan^2(x)-\tan(x)+x}2+C$$

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.

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

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?

• Thanks for your reply 😀 – Henry Trancer Oct 25 '18 at 22:10

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 ?

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 $$u = \tan(x)$$.