integrate problem

**jvignacio** · November 1st 2008, 09:56 PM

hi just confused on this integration.

$<br /> \int xcosx<br />$

how would i go about solving?

**mr fantastic** · November 1st 2008, 10:00 PM

Originally Posted by jvignacio

hi just confused on this integration.

$<br /> \int xcosx<br />$

how would i go about solving?

There are many techniques. One such is integration by parts. You'll find it done as an example somewhere in here: http://en.wikipedia.org/wiki/Integra...#More_examples.

**jvignacio** · November 1st 2008, 10:10 PM

Originally Posted by mr fantastic

There are many techniques. One such is integration by parts. You'll find it done as an example somewhere in here: Integration by parts - Wikipedia, the free encyclopedia.

whats the other way besides integration by parts? coz we dont learn that technique in my course. Its in the next course up from mine.

**mr fantastic** · November 1st 2008, 10:13 PM

Originally Posted by jvignacio

whats the other way besides integration by parts? coz we dont learn that technique in my course. Its in the next course up from mine.

Try integration by recognition:

Differentiate y = x sin x.

Hence find the integral of x cos x.

**jvignacio** · November 1st 2008, 10:21 PM

Originally Posted by mr fantastic

Try integration by recognition:

Differentiate y = x sin x.

Hence find the integral of x cos x.

well thats exactly what the tutorial question was . find diff of xsinx and then using that answer integrate xcosx...

i found the diff of xsinx.. it was sinx + xcosx but im not sure how this can lead me to the integral of xcosx

**mr fantastic** · November 1st 2008, 10:30 PM

Originally Posted by jvignacio

well thats exactly what the tutorial question was . find diff of xsinx and then using that answer integrate xcosx...

i found the diff of xsinx.. it was sinx + xcosx but im not sure how this can lead me to the integral of xcosx

$\frac{dy}{dx} = \sin x + x \cos x$

Integrate both sides with respect to x:

$y = \int \sin x + x \cos x \, dx = \int \sin x \, dx + \int x \cos x \, dx = - \cos x + \int x \cos x \, dx$

Substitute $y = x \sin x$ :

$x \sin x = - \cos x + \int x \cos x \, dx$

Now re-arrange to make $\int x \cos x \, dx$ the subject and add a "+ C".

Next time post the whole question - as you can now see the first part was essential to answering the second part.

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