Evaluating this integral $ \small\int \frac {x^2 dx} {(x\sin x+\cos x)^2} $

Question

The question:

Compute$$ \int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2} $$

Tried integration by parts. That didn't work.

How do I proceed?

Well, it is there in the problem sheet that we are working at. (Btw, what makes you think it DOESN'T have an anti-derivative?) — Parth Thakkar, Apr 19 '13 at 14:48
@ParthThakkar: Given a random function, it most likely does not have an anti-derivative in terms of elementary functions. of course, if you tell the source, one would not have had to ask you this question. — Aryabhata, Apr 19 '13 at 15:01
Ok, got it! Now onwards, such things will be mentioned if required. — Parth Thakkar, Apr 19 '13 at 15:03
@Vijay: Now I suspect there is a typo. Are you sure the denominator is not $(x \sin x + \cos x)^2$? (In which case lab's answer would be perfect). — Aryabhata, Apr 19 '13 at 15:22
Yes, sorry for the error. Editing. (also scolding Vijay for this XD) — Parth Thakkar, Apr 19 '13 at 15:27
@ParthThakkar, lol, I'm somewhat sure that i posted the right question — dajoker, Apr 19 '13 at 15:35
@VijayRaghavan: my apologies if this caused an error, there is yet no mechanisms to lock things when people are working on them and when I was trying to review and fix the edits, things seemed to have gone astray. Regards — Amzoti, Apr 19 '13 at 15:38

lab bhattacharjee · Accepted Answer · 2013-04-19T15:38:17.823

19

$$\text{Observe that, }\frac{d(x\sin x+\cos x)}{dx}=x\cos x$$

$$ \int \frac {x^2 \, \operatorname{d}\!x} {(x\sin x+\cos x)^2} =\int \frac x{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x)^2}dx$$

So, if $z=x\sin x+\cos x, dz=x\cos xdx$

So, $\int \frac{x\cos x}{(x\sin x+\sin x)^2}dx=\int \frac{dz}{z^2}=-\frac1z=-\frac1{x\sin x+\cos x}$

So, $$I=\frac x{\cos x}\int \frac{x\cos x}{(x\sin x+\cos x)^2}dx-\int \left(\frac{d(\frac x{\cos x})}{dx}\int \frac{x\cos x}{(x\sin x+\cos x)^2}dx\right)dx$$

$$=-\frac x{\cos x(x\sin x+\cos x)}+\int \left(\frac{x\sin x+\cos x}{\cos^2x}\right)\left(\frac1{x\sin x+\cos x} \right)dx$$

$$=-\frac x{\cos x(x\sin x+\cos x)}+\int\sec^2xdx$$

$$=-\frac x{\cos x(x\sin x+\cos x)}+\tan x+C$$ where $C$ is an arbitrary constant of indefinite integral

$$\text{Another form will be } \frac{\sin x-x\cos x}{x\sin x+\cos x}+C$$

edited Apr 19 '13 at 15:38

answered Apr 19 '13 at 14:56

lab bhattacharjee

274,582

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Damn! We had observed the derivative thingy, but couldn't think of multiplying and diving by $cos x$. Smart! – Parth Thakkar Apr 19 '13 at 14:58
Thanks.. Got the solution. – dajoker Apr 19 '13 at 14:59
@VijayRaghavan, my pleasure. – lab bhattacharjee Apr 19 '13 at 15:04
@Aryabhata, is it ok now? – lab bhattacharjee Apr 19 '13 at 15:14
Same here can't understand :( – ABC Apr 19 '13 at 15:21
@Aryabhata, let me rectify, misread the Question. Thanks for your input – lab bhattacharjee Apr 19 '13 at 15:22
1

What?????????? It was a bigggggg mistake. Haha . Even the answer was accepted and 3 up-votes. @VijayRaghavan and others all trolled..... :D – ABC Apr 19 '13 at 15:25
@exploringnet, if it was big enough, how it eluded both the acceptor & the solver? – lab bhattacharjee Apr 19 '13 at 15:28
Aah! now the question is edited. – ABC Apr 19 '13 at 15:30
How can you take the $x/\cos x$ out of the integral????? – marty cohen Oct 01 '14 at 17:49
2

@martycohen, I have used Integration by Parts(http://en.wikipedia.org/wiki/Integration_by_parts), $$\int u(x)v'(x) dx=u(x)v(x)-\int[u'(x)v(x)]dx$$ Setting $v'(x)=w(x)\implies v(x)=\int w(x)dx$ to find $$\int u(x)w(x)dx=u(x)\int w(x)dx-\int[u'(x)\cdot\int w(x)dx]dx$$ – lab bhattacharjee Oct 01 '14 at 18:12
@labbhattacharjee what is u and v here – Koolman Jul 13 '16 at 18:51
@labbhattacharjee can you help me in http://math.stackexchange.com/q/1857096/298843 – Koolman Jul 13 '16 at 19:45

score 17 · Answer 2 · edited Aug 31 '16 at 05:01

17

Differentiation of $$x\sin x+\cos x \space \text {is}\space x\cos x $$ \begin{align}\int\underbrace {\frac{x \cos x}{(x\sin x+\cos x)^2}}_{\text {II}}\cdot\underbrace {\frac {x}{(\cos x)}}_{\text {I}}{d}x\end{align} Now integrate by parts. $$I=\frac{-1}{(x\sin x+\cos x)}.\frac {x}{(\cos x)} +\int\frac{1}{(x\sin x+\cos x)}.\frac{\cos x.1 -x(-\sin x)}{\cos^2 x} $$ Now, I hope things are clear to you.

edited Aug 31 '16 at 05:01

Martin Sleziak

53,687

answered Aug 21 '16 at 13:03

Aakash Kumar

3,480

Will it not be too long – Koolman Aug 21 '16 at 15:26
@koolman No.., Give it a try . – Aakash Kumar Aug 21 '16 at 15:28
in first step will we take $d(xsin x + cos x)$ – Koolman Aug 21 '16 at 15:30
@koolman Yes..${}{}{}{}$ – Aakash Kumar Aug 21 '16 at 15:38
then it will be tooo long..... – Koolman Aug 21 '16 at 15:39
I think you have a typo: $\cos x\cdot 1-x (-\sin x) $ as numerator of the second fraction in the last integrand. – Américo Tavares Aug 21 '16 at 17:41
1

@AméricoTavares Thanks .. – Aakash Kumar Aug 21 '16 at 17:45

score 16 · Answer 3 · edited Apr 13 '17 at 12:21

16

I was inspired by this post to conduct this method.

$$\int\frac{x^2}{(x\sin x+\cos x)^2}\mathrm{d}x $$

The Harmonic Addition Theorem comes in handy, so $x \sin x + \cos x = \sqrt{1+x^2} \cos(x-\alpha)$ where $\alpha = \arctan(x)$.

The origin integration becomes:

$$\int \frac{x^2}{1+x^2} \sec^2(x-\alpha) \mathrm{d}x $$

Notice that $\int \frac{x^2}{1+x^2} dx = x - \arctan x = x-\alpha $.

So let $t = x-\alpha$ then $dt = \frac{x^2}{1+x^2} dx$ and the integration simplifies as:

$$ \int \sec^2(t) dt = \tan(t) = \tan(x - \arctan x)$$.

In conclusion, $$\int\frac{x^2}{(x\sin x+\cos x)^2}\mathrm{d}x =\tan(x - \arctan x)+C$$

edited Apr 13 '17 at 12:21

Community

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answered Aug 21 '16 at 13:15

Zau

3,909
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Is that $a = \arctan(x)$ meant to be $\alpha = \arctan(x)$? – someonewithpc Aug 21 '16 at 23:33
@someonewithpc Yes a typo. – Zau Aug 21 '16 at 23:36

score 2 · Answer 4 · edited Apr 17 '23 at 17:06

2

$$\int\frac{x^2}{(x\sin x+\cos x)^2}\mathrm{d}x = \int (x\sec x)\frac{x \cos x}{(x\sin x+\cos x)^2}\mathrm{d}x$$ Take $xsecx$ as first functions and apply integreation by parts. there you go boom!!

edited Apr 17 '23 at 17:06

Shub

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answered Feb 05 '17 at 13:09

Pushkar Soni

163
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Evaluating this integral $ \small\int \frac {x^2 dx} {(x\sin x+\cos x)^2} $

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