So I am a beginner to integral calculus, and I am trying to solve this question which is confusing me
It is $$\int \frac{\arcsin{x}}{(1-x^2)^{\frac{3}{4}}}dx$$
So far I have tried using substitution of $t=\arcsin{x} \Rightarrow dt=\frac{dx}{(1-x^2)^{\frac{1}{2}}}$. I can rewrite the denominator as $(1-x^2)^{\frac{1}{2}} (1-x^2)^{\frac{1}{4}}$ so that I can put in $dt$
And then I substituted $x=\sin{t}$ (so $1-x^2=\cos^2{t}$, which I can put in $(1-x^2)^{\frac{1}{4}}$ as $\sqrt{\cos{t}}$) back into the integral and it
left me with this
$$\int \frac{t}{\sqrt{\cos{t}}}dt$$
I don't know how to proceed from here, or whether I'm approaching this problem wrong.
Could someone tell me how to go about this problem and how to think about the strategy?
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asterbot
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Does this answer your question? What is the integral of $\frac{\arcsin x}{(1-x^2)^{3/2}}$? – Jan 30 '22 at 18:35