Question

how to integrate x^2sinxdx by parts?

Answer 1

uv formulae of integration \[intuv=u \int\limits(v)+v \int\limits(u)\]

Answer 2

that one will work well with a tabular method

Answer 3

x^2 (-cosx)+ 2x*sinx

Answer 4

If I may rephrase that :)

\[\int\limits_{}^{}f(x)g'(x)dx = f(x)g(x) - \int\limits_{}^{}f'(x)g(x)dx\]

Shall I show you the steps? :)

Answer 5

@terenzreignz yes please

Answer 6

So we use the formula posted above
We let
x^2 = f(x)
sin x = g'(x)

We need to find f'(x) and g(x)
f'(x) = 2x
g(x) = -cos x (since g'(x) = -sin x)

So, plugging everything in...
\[\int\limits_{}^{}x^{2}\sin x dx = -x^{2}cosx - \int\limits_{}^{}(-2x \cos x)dx\]

Answer 7

And it gives us a new integral, we just have to integrate that by parts as well.
Are you getting me so far? :)

Answer 8

@terenzreignz - is there actually a product rule for integration like rajesh says?? :O

Answer 9

This is pretty much it :|
It's a shame integration isn't as straighforward as differentiation :(
You have to go through a lot of uhh, tedious work to integrate products like these :)

I can derive the formula for integration by parts, if any of you are interested :)

Answer 10

ok

Answer 11

so the answer for the previous question

Answer 12

So we have now
\[-x^{2} \cos x - \int\limits_{}^{}(-2x \cos x)dx\]
We integrate by parts again, but first, let's simplify a bit.
\[-x^{2} \cos x + \int\limits_{}^{}2x \cos xdx\]
Ok, so now we focus on this integral.
We let
f(x) = 2x and
g'(x) = cos x
THEN

f'(x) = 2
g(x) = sin x (since g'(x) = cos x)

So our equation becomes:
\[-x^{2}\cos x + (2x \sin x - \int\limits_{}^{}2 \sin x dx)\]
I have complete faith that you can evaluate that last integral :)
If not, feel free to ask me again :D
Cheers :)

Answer 13

While you're perusing that, here's how to derive the formula for integration by parts :D
We start with the product rule for differentiation:
\[\frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x)\]
We multiply the differential dx to both sides of the equation:
\[df(x)g(x) = f(x)g'(x)dx + f'(x)g(x)dx\]
And now we integrate both sides:
\[\int\limits_{}^{}df(x)g(x) = \int\limits_{}^{}f(x)g'(x)dx + \int\limits_{}^{}f'(x)g(x)dx\]
\[f(x)g(x) = \int\limits_{}^{}f(x)g'(x)dx + \int\limits_{}^{}f'(x)g(x)dx\]
And then we just manipulate this equation to give us:
\[\int\limits_{}^{}f(x)g'(x)dx = f(x)g(x) - \int\limits_{}^{}f'(x)g(x)dx\]

This actually is still a bit cumbersome, let's 'tame' it a bit
let
u = f(x)
v = g(x)
du = f'(x)dx
dv = g'(x)dx

So that gives us a neater :
\[\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu\]
Enjoy your calculus! :D

Answer 14

@terenzreignz AMAZINGLY DONE!!! WOW!!! you smart pal! you must have great interest in it don't ya? btw can you evaluate the last integral, wanna check my answer with the integral

Answer 15

Thanks, I'd love to check your answer :D
and thanks again :D

Answer 16

2cosx+2xsinx-x^2cosx+c correct?

Answer 17

That's right!
Oh, but if you're not sure, you can always try differentiating it, see if you end up with the original expression :)