How do I solve the integral:

$$\int\limits_{0}^{1}\sqrt{1-x^{2}}x^2dx$$

Thanks in advance!

Question

How do I solve the integral:

$$\int\limits_{0}^{1}\sqrt{1-x^{2}}x^2dx$$

Thanks in advance!

amistre64 · Answer

$$\int\limits_{0}^{1}\sqrt{1-x^{2}}x^2dx$$

amistre64 · Answer

... now it formats ..

anonymous · Answer

Thanks

amistre64 · Answer

what methods have you tried? or know of?

anonymous · Answer

looks like a trig sub to me

anonymous · Answer

good morning sensei!

amistre64 · Answer

i was considering a u sub ...

amistre64 · Answer

Howdy sate

anonymous · Answer

Well I suppose I could try partial integration, I have a formula for integrals of the form 

$$\int\limits_{}^{}\sqrt{a-x^2}$$

I just think it is to complicated to solve. I hoped for a good substitution!

anonymous · Answer

try 
$$x=\sin(u)$$ and then you will get powers of sine to integrate. use reduction formula to finish it

anonymous · Answer

trust me it is set up for trig sub.

$$u=\sin(x)$$
$$du = \cos(x)dx$$ 
$$\int \cos(u) \sin^2(u)\cos(u)du $$
$$\int \sin^2(u)\cos^2(u)du $$

amistre64 · Answer

just taking a stab at it ....

spose u = 1-x^2, then du = -2x dx such that dx = du/-2x
$$\int \cfrac{\sqrt{u}x^2}{-2x}du$$
$$\int \cfrac{\sqrt{u}x}{-2}du$$
but x = sqrt(1-u) soo
$$\int \cfrac{\sqrt{u}\sqrt{1-u}}{-2}du$$
$$\int \cfrac{\sqrt{u}\sqrt{1-u}}{-2}du$$ ... prolly not gonna play nice this way is it :)

amistre64 · Answer

trig it is ;) good call sate

anonymous · Answer

then one of those damn reduction formulas i forget.  i can look in a calc book because 
$$\int\sin^2(u)\cos^2(u)du$$ is probably even an example.

anonymous · Answer

Hey thanks! I solved it by satellite's substitution x=sinu! 

The result then is integral sin^2(u) cos^2(u) du = 1/32 (4 u-sin(4 u))+constant

anonymous · Answer

Thanks for the effort!

anonymous · Answer

good work.  glad you did it and not me. i always forget how to do powers of sine and cosine. you have to use some reduction formula

anonymous · Answer

ok ajahangir but of course now you have to go back, yes?  
$$u=\sin^{-1}(x)$$ and so on

anonymous · Answer

Oh, I always forget that step.. Thanks, suppose the answer will not look as pretty anymore

anonymous · Answer

Actually, I do not have to go back if I transform the boundaries as well, right? If the boundaries are 1 and 0 for x, then they are (pi)/2 and 0 for x=sinu, right?

This would give the result: pi/16 !

anonymous · Answer

that is a better idea than going back, and your answer is right