I was solving the integration of inverse trigonometric function and faced a question which i find it hard to understand. I need to find the definite integration of this function.

$$\int_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x^2}} dx$$

I tried to use the substitutional method by

$$u= \sin^{-1}(x)$$

and getting $\dfrac{du}{dx}= \dfrac{1}{\sqrt{1-x^2}}$

and $dx= du(\sqrt{1-x^2})$

but when i substitute that into the function, it does not make any sense. This is where i got stuck (not even sure if i did in the right way or not..)

am i doing it right? Should I use another method to approach to the answer? (Sorry if this question is duplicating, i could not find an appropriate answer..)

  • $\begingroup$ The substitution is good. We have $\frac{dx}{\sqrt{1-x^2}}=du$. So you end up needing $\int u\,du$. $\endgroup$ – André Nicolas May 19 '14 at 17:06
  • 2
    $\begingroup$ What are you stuck with? Your substitution is good, just calculate $\int_0^{\pi/6} u\, du$ $\endgroup$ – gar May 19 '14 at 17:09
  • $\begingroup$ Oh my god, so it was the right way.. i figured that i made a simple mistake during the substituting which made the whole function weird. Thanks a lot for your helps!! $\endgroup$ – Sam May 19 '14 at 17:15

$$ \int_0^{1/2} \underbrace{(\sin^{-1} x)}_{u} \underbrace{\left( \frac{dx}{\sqrt{1-x^2}} \right)}_{du} = \int u\,du $$

When $x=0$ then $u=0$ and when $x=1/2$ then $u=\pi/6$, so you actually get $$ \int_0^{\pi/6} u\,du. $$

  • $\begingroup$ Thanks for your help! I managed to solve it :) $\endgroup$ – Sam May 19 '14 at 17:19

Let $y=\arcsin x$, then $x=\sin y\;\Rightarrow\;dx=\cos y\ dy=\sqrt{1-x^2}\ dy$, then $$ \begin{align} \require{cancel}\int_{x=0}^{\Large\frac12}\frac{\arcsin x}{\sqrt{1-x^2}}\ dx&=\int_{x=0}^{\Large\frac12}\frac{y}{\cancel{\sqrt{1-x^2}}}\cdot\cancel{\sqrt{1-x^2}}\ dy\\ &=\int_{x=0}^{\Large\frac12} y\ dy\\ &=\left.\frac12y^2\right|_{x=0}^{\Large\frac12}\\ &=\frac12\arcsin^2\left(\frac12\right)-\frac12\arcsin^2\left(0\right)\\ &=\large\color{blue}{\frac{\pi^2}{72}}. \end{align} $$


Notice the derivative of $\mathrm{Arcsin} \,x$ is $\frac{1}{\sqrt{1-x^2}}$, hence your integrand is of the form $u u'$ with $u=\mathrm{Arcsin}\,x$, and a primitive is $u^2/2$. Thus

$$\int_{0}^{\frac{1}{2}}\frac{\mathrm{Arcsin}\, x}{\sqrt{1-x^2}} dx= \frac12\left[\mathrm{Arcsin}^2\,x\right]_0^{1/2}=\frac12\mathrm{Arcsin}^2\,\frac12=\frac{\pi^2}{72}$$

  • $\begingroup$ Why do you capitalise the 'a'? $\endgroup$ – user85798 May 26 '14 at 3:35
  • $\begingroup$ @Oliver It's the way the function is "traditionally" written in french. But ISO 31-11 recommends lowercase for the inverse trig functions... $\endgroup$ – Jean-Claude Arbaut May 26 '14 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.