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If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.

My solution:

$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$ Then let $x^3=t\implies 3x^2\cdot dx=dt$

$$\implies\dfrac{1}{3}\int\dfrac{1}{x}\cdot\dfrac{3x^2}{x^2}\sqrt{\dfrac{x^3}{a^3-x^3}}dx\\=\dfrac{1}{3}\int\dfrac{1}{t}\sqrt{\dfrac{t}{a^3-t}}dt\\ =\dfrac{1}{3}\int{\dfrac{1}{\sqrt{ta^3-t^2}}}dt\\=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{\frac{a^6}{4}-\bigg(t-\frac{a^3}{2}\bigg)^2}}}dt\\=\dfrac{1}{3}\sin^{-1}\bigg(\dfrac{2t-a^3}{a^3}\bigg)+C$$ $$=\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C\tag{1}$$

This is not of desired form but when I drew graph of: $\color{grey}{1.57+\sin^{-1}\left(2x^2-1\right)}$ and $\color{green}{2\sin^{-1}x}$, they coincides for $\color{red}{x>0}$. Green graph is for $\ 2\sin^{-1}x\ $. enter image description here

So from $(1)$ $$\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C=\dfrac{2}{3}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$

If I didn't made any mistake then please help me getting $1.57+\sin^{-1}\left(2x^2-1\right)\equiv\ 2\sin^{-1}x\ $ for $x>0$ without any graphs.

Also, I found that if I substitute $x^3=a^3\sin^{2}x$ then we reach the desired form of the result directly. But I'm an enthusiast to continue my previous solution. Please help.

Thanks!

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    $\begingroup$ I would personally start with the right side and differentiate instead of integrating the left side. $\endgroup$ – Arthur Jul 4 '18 at 6:47
  • $\begingroup$ @Arthur I'm just showing loyalty :) $\endgroup$ – mnulb Jul 4 '18 at 6:57
  • $\begingroup$ I don't understand this downvote! I was just asking why there are some limits to carry forward my work to reach the desired form. Is that the limit of Mathematics or human minds? People are answering not do this question in this way, rather differentiate it, this doesn't answer my problem. Moreover, I thought I found an answer to my problem, as when I differentiate both expressions i.e $2\sin^{-1}x$ and $\sin^{-1}(2x^2-1)$, I got the same slope for $x>0$, which let me conclude that if I shift these function along $y$-axis, they'll coincide once. I'll appreciate your take on my proof. $\endgroup$ – mnulb Jul 4 '18 at 9:34
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Since you already have the form you must get the answer into then why not get some hints from it.

$$\int \frac {\sqrt x}{\sqrt {a^3-x^3}} dx=\int \frac {\sqrt x}{a^{\frac 32}\sqrt {1-\left(\frac {x^{3/2}}{a^{3/2}}\right)^2 }} dx$$

Let $$\frac {x^{3/2}}{a^{3/2}}=t$$ hence $$dt=\frac 32\cdot \frac {\sqrt x}{a^{3/2}}dx$$

Hence the integral changes to $$\int \frac 23 \cdot \frac {1}{\sqrt {1-t^2}} dt $$

Hence the answer $b=3,d=2$ follows

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Hint: As suggested in the comments, differentiate both sides of the equation.

Edit: If you do this, you obtain $$\left(\frac{x}{a^3-x^3}\right)^{1/2}=\frac db \frac {1}{\sqrt{1-[\left(\frac xa\right)^{3/2}]^2}}\cdot \left(\frac1 a\right)^{3/2}\frac32 x^{1/2}.$$ Hopefully, you can rearrange this to obtain $$\frac{x^{1/2}}{(a^3-x^3)^{1/2}}=\frac{3d}{2b}\cdot \frac{x^{1/2}}{(a^3-x^3)^{1/2}},$$ which implies $$\frac db=\frac23,$$ so that $d=2$ and $b=3$ since $d$ and $b$ are coprime.

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