# Integration by parts

1. Feb 26, 2009

### apples

I am stuck with 2 questions.

1. http://132.239.150.164/math/img/math_967.5_fabcf8ac5b3f2459050993af33426844.png[/URL]

2. http://132.239.150.164/math/img/math_972_41ec111060693034ac0775fd52778392.png[/URL]
2. Relevant equations

http://132.239.150.164/math/img/math_989.5_89c6805f9e6a047e24c9d1b535209efd.png

3. The attempt at a solution
1.
I take u= (4-x)^(1/2)
dv = xdx
du = 1/2* (4-x)^(-1/2)
v= (x^2)/2
I get stuck at

((x^2)/2)*(4-x)^(1/2)- ∫[(x^2)dx/4*(4-x)^(1/2)

2. I take u= (arcsin x)^2 dv=dx or 1
du= (2arcsin x)/((1-x^2)^(1/2)]
v=x

so

x(arcsin x)^2 -2 ∫[(x arcsin x dx)/(1-x^2)^(1/2)]

I don't know what to do after that.

How do I solve the red.

Last edited by a moderator: Apr 24, 2017
2. Feb 26, 2009

### gabbagabbahey

Try $u=x$ and $dv=\sqrt{4-x}dx$ instead

Try by parts once more, this time use $u=\arcsin(x)$ and $dv=\frac{x}{\sqrt{1-x^2}}dx$

3. Feb 27, 2009

### Staff: Mentor

A much simpler approach is to let u = 4 - x, so du = -dx.
The indefinite integral becomes
$$\int -(4 - u)u^{1/2}du = -\int (4u^{1/2} - u^{3/2})du$$

After you have an antiderivative, undo the substitution and use your limits of integration. You should always check to see if an ordinary substitution will get the job done before bringing in the big guns, such as integration by parts.

Last edited by a moderator: Apr 24, 2017