Integral (xsin^3(x^2)(cosx^2))dx?

Answer 1

Observe that d/dx sin(x^2) = cos(x^2) 2x

So, your integral can be written as

1/2 Integral sin^3(x^2) 2x cos(x^2) dx = 1/2 Integral sin^3(x^2) d/dx (sin(x^2) dx , an integral of the type Int u^m du.

So, we get ^1/2 * 1/4 sin^4(x^2) + C = (1/8) sin^4(x^2) + C

Answer 2

Using the method I explained here
http://answers.yahoo.com/question/index;_ylt=AkAUYp9EYOSIB22oTy4EgBDsy6IX;_ylv=3?qid=20071127140723AAe8sov&show=7#profile-info-mEUJd7hvaa

If you differentiate sin x you get cos x. So increase the sinx terms power in the equation above by 1 and divide by the same power.

So Ans: (Sin^4 x^2) / (4 * 2)

I got the * 2 in the denominator so that when you differentiate the lot, when diff x^2 you get 2x, so adding * 2 in denominator will cancel it out.

When Integrating something always differentiate your ans so that you get back to the question. By this way I spotted the necessity of * 2 in the denominator to compensate the 2 which you get from 2x. (Use chain rule for differentiating)

Hope this helps.

Answer 3

∫x.sin³(x²)cos(x²)dx

Let u = x²
du = 2xdx

∫x.sin³(x²)cos(x²)dx
= ∫(1/2)sin³(u)cos(u)du

v = sin(u)
dv = cos(u)du

∫(1/2)sin³(u)cos(u)du
= ∫(1/2)v³dv
= (v^4)/8 + c
= (1/8)sin^4(u) + c
= (1/8)sin^4(x²) + c