How do you evaluate #int arcsinx/sqrt(1-x^2)# from #[0, 1/sqrt2]#?

2 Answers
Noah G · mason m
Feb 24, 2017

#pi^2/32#

Explanation:

Note that #d/dxarcsinx = 1/sqrt(1 - x^2)#. Therefore, use the substitution #u = arcsinx#. This means that #du= 1/sqrt(1 - x^2)dx# and #dx= sqrt(1 - x^2)du#. Adjusting the bounds of integration accordingly, we have:

#int_0^(pi/4) u/sqrt(1 - x^2) sqrt(1 - x^2)du#

#int_0^(pi/4) u du#

#[1/2u^2]_0^(pi/4)#

We don't have to reverse the substitution, because we adjusted the bounds of integration. If you evaluated in #u# with the initial bounds of integration, the answer would be incorrect.

#1/2(pi/4)^2#

#pi^2/32#

Hopefully this helps!

Monzur R.
Feb 25, 2017

This is another, and, in my opinion, simpler method to evaluate the integral. However, the answer below is totally valid, and if it helps, use that method instead of this one.

Explanation:

When it comes to integrating this function, one may be tempted to use #u#-substitution and change the variables in order to simplify the integrand.

But knowing that #d/dxarcsinx=1/(sqrt(1-x^2))#, we know that our integral #int_0^(sqrt2/2)arcsinx/(sqrt(1-x^2)) dx# is in the form #intf(x)f'(x)dx#. And the reverse chain rule states that this can be evaluated as #1/(n+1)[f(x)]^(n+1)+"C"#.

So we can evaluate our integral as #[1/2arcsin^2x]_0^(sqrt2/2)=1/2arcsin^2(sqrt2/2)-1/2arcsin^2(0)=pi^2/32#

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