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

$$2\int x^2\, \sec^2x \,\tan x\, \mathrm{d}x$$

How to solve this using integration by parts? WolframAlpha can solve it, but is unable to give a step-by-step solution, and has a different answer to the one in the back of my textbook. There is also a question/answer on yahoo answers, but yet again, that gives a different answer to the one given to me.

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Have u considered Integration by parts? –  Gautam Shenoy Apr 12 '13 at 12:13
I have, but I am unable to integrate $sec^2 x \tan x$ –  LordAro Apr 12 '13 at 12:15
@LordAro: See the answer and how to choose $dv$ for the first integration by parts. –  Mhenni Benghorbal Apr 12 '13 at 12:24

Integrate by parts, differentiating $x^2$ and integrating $\sec ^{2}x\tan x$ by substitution$^1$:

$$\begin{eqnarray*} I &=&\int x^{2}\sec ^{2}x\tan x\, dx=\frac{1}{2}x^{2}\sec ^{2}x-\int x\sec ^{2}x\,dx \\ &=&\frac{1}{2}x^{2}\sec ^{2}x-\left( x\tan x-\int \frac{\sin x }{\cos x}\,dx\right)\qquad \text{by parts; note }^2 \\ &=&\frac{1}{2}x^{2}\sec ^{2}x-x\tan x-\ln |\cos x|+C. \end{eqnarray*}$$ So $$\begin{equation*} 2I=2\int x^{2}\sec ^{2}x\tan x dx=x^{2}\sec ^{2}x-2x\tan x-2\ln | \cos x|+C. \end{equation*}$$

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$^1$ Let $u=\sec x$. Then $du=\sec x\, \tan x$ and

$$\int \sec ^{2}x\tan xdx=\int u\,du=\frac{1}{2}u^{2}=\frac{1}{2}\sec ^{2}x.$$

$^2$ Differentiate $x$ and integrate $\sec^2 x$. Since $\dfrac{d}{dx}\tan x=1+\tan ^{2}x=\sec ^{2}x$, $\displaystyle\int \sec ^{2}xdx=\tan x$.

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That looks good, but can you expand on the $\int sec^2 x tan x dx$ ? I'm not really sure where the $1/2 sec^2 x$ comes from –  LordAro Apr 12 '13 at 13:00
@LordAro We can integrate it by the substitution $u=\sec x$ (see edit). –  Américo Tavares Apr 12 '13 at 13:25
great, thanks :) I think i'm going to have to revise the substitution method –  LordAro Apr 12 '13 at 13:36
@LordAro You are welcome. –  Américo Tavares Apr 12 '13 at 13:40
Hint: Integrate by parts twice and in each case assume $u$ to be the polynomial function and consider $dv$ for the first integration by parts as $$dv = \sec(x) (\sec(x)\tan(x)) dx \implies v = \frac{1}{2} \sec^2(x).$$