<s>[INST] Evaluating indefinite integrals. 
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\ $.  
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!
 [/INST] 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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