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# Calc integration help?

integral of xsinx^2(cosx^2)^8

### 2 Answers

- husoskiLv 73 weeks ago
Look inside the outermost "messy part" of the integrand for a possible substitution.

Inside of (cos x²)^8 is cos x², so try that. Let u = cos x², and the chain rule gives:

du/dx = -2x sin x²

That will work, making (cos x²)^8 = u^8 and x sin x² dx = - (1/2) du. So:

∫ x sin x² (cos x²)^8 dx = -(1/2) ∫ u^8 du = (-1/18) u^9 + C = (-1/18) (cos x²)^9 + C

- 3 weeks ago
When integrating, always include the differential. It's important

x * sin(x^2) * cos(x^2)^8 * dx

Use a substitution

a = x^2

da = 2x * dx

x * sin(x^2) * cos(x^2)^8 * dx =>

(1/2) * sin(x^2) * cos(x^2)^8 * 2x * dx =>

(1/2) * sin(a) * cos(a)^8 * da =>

(1/2) * cos(a)^8 * sin(a) * da

One more substitution

u = cos(a)

du = -sin(a) * da

(1/2) * cos(a)^8 * sin(a) * da =>

(-1/2) * cos(a)^8 * (-sin(a) * da) =>

(-1/2) * u^8 * du

Now we can integrate.

(-1/2) * (1/9) * u^9 + C =>

(-1/18) * cos(a)^9 + C =>

(-1/18) * cos(x^2)^9 + C

There you go.

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@Captain: Thank you!