Integration by Parts: Solving x^2(cosx)dx with Step-by-Step Guide

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Homework Statement


solve by using intergration by parts x^2(cosx)dx

Homework Equations


My question is i got this answer but my computer algebra system gave me this answer
(x^2-2)sin(x)+2xcos(x) can you tell me where i went wrong??

The Attempt at a Solution


u=X^2
dv=cosx
du=dx
v=sinx

Sx^2(cosx)dx=uv-Svdu=x^2(sinx)-Ssinxdx

S2xcosxdx=2xsinx-(-cosx)+c= final answer= 2xsinx+cosx+C
 
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If u=x^2, du=2x*dx. Not du=dx. You have to do integration by parts twice. And why are you integrating 2xcos(x)?
 
oo thank you
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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