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Find $$\int \frac {1}{\sqrt{x(1-x)}}\,dx$$ using the the substitution $x=\sin^2(u)$

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    what are your efforts? – 2017-02-05
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    So, when you use the substitution, as suggested, what does $dx$ equal? What is the integrand after substitution? This is child's play. If I have an integral $\int (x+1)\,dx$, and was told to substitute $x+1 = u$, then $dx = du$, and then fill in the blanks $$\int (\underbrace{\underline{\;\;\;\;}}_{x+1 = u})\; (\underbrace{\underline{\;\;\;}}_{dx = du})$$, what is the result? Surprise! $$\int u\,du$$ – 2017-02-05

2 Answers 2

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Let $x = \sin^{2}(u)$ then $dx = 2\sin(u)\cos(u)\,du$, so the integral becomes: \begin{align*} \int \dfrac{2\sin(u)\cos(u)du}{\sqrt{\sin^2(u)(1-\sin^2(u))}} &= 2 \int \dfrac{\sin(u)\cos(u)\,du}{\sqrt{\sin^{2}(u)\cos^{2}(u)}} \\ &= 2 \int \dfrac{\sin(u)\cos(u)du}{\sin(u)\cos(u)} \\ &= 2 \int \,du \\ &= 2u +C \\ &= 2\arcsin(\sqrt{x}) + C \end{align*}

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    I think there is a mistake with your last step. $x= \sin^2(u)$ doesn't mean $u=\sin^2(x)$ .. – 2017-02-05
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    Yeah, good catch. – 2017-02-05
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    The substitution should be, more properly, $u=\arcsin\sqrt{x}$, so it's clear that $0$\sin u$ and $\cos u$ are positive in the stated interval. – 2017-02-05
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    Should it also be + or - √x ? – 2017-02-05
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Hint

With the substitution $x= \sin^2(u),$ we have $dx = 2 \sin(u) \cos(u) \ du$ Figure out the transformed integrand.

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    Would the cowards who are down voting care to identify themselves? – 2017-02-05
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    I don't understand why I got downvoted.. I just offered a helpful hint... – 2017-02-05
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    I did too. Here (+1) to helo – 2017-02-05