<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] 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
</s>