Integration by parts with √(4-x) and arcsin²x

[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]

Homework Equations

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

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.

 

[gabbagabbahey]

The Attempt at a Solution

1.
I take u= (4-x)^(1/2)
dv = xdx

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

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.

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

 

[Mark44]

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]

Homework Equations

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

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)

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.