Saturday

August 30, 2014

August 30, 2014

Posted by **Anonymous** on Wednesday, November 7, 2012 at 11:58am.

- Maths - Integration by parts -
**Steve**, Wednesday, November 7, 2012 at 2:13pmLet I = ∫x^2 sinx dx

u = x^2

du = 2x dx

dv = sinx

v = -cosx

∫u dv = uv - ∫v du

= -x^2 cosx + ∫2x cosx dx

now, for ∫x cosx dx,

u = x

du = dx

dv = cosx dx

v = sinx

I = -x^2 cosx + 2(x sinx - ∫sinx dx)

= -x^2 cosx + 2x sinx + 2cosx + C

= 2x sinx + (2-x^2)cosx + C

**Related Questions**

Integral - That's the same as the integral of sin^2 x dx. Use integration by ...

math - I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up ...

integration by parts - s- integral s ln (2x+1)dx ? = ln(2x+1)x - s x d( ln (2x+1...

Calculus II - Evaluate the integral using method of integration by parts: (...

Math/Calculus - How would I integrate the following by parts: Integral of: (x^2...

Calculus Problem - I need to find the integral of e^(2x)sin(3x) I used ...

calc - evaluate the integral: y lny dy i know it's integration by parts but i ...

trig integration - s- integral endpoints are 0 and pi/2 i need to find the ...

maths - use an iterated integral to find area of region bounded by graphs sin(x...

double integral - 1. Sketch the region of integration & reverse the order of ...