I was attempting this integral: $I = \int \dfrac{\arcsin x}{(1-x^2)^{3/2}} dx$. But I am getting the wrong answer.
My approach:
Let $\arcsin x = t \implies dx = \sqrt{1-x^2} dt$. So, $I =\frac{t}{\cos^2 t} dt = t × \tan t - \int \tan t dt = \dfrac{\arcsin x × x}{\sqrt {1-x^2}} + \ln (\cos t) + K $.
But apparently, the correct answer includes a negative sign in the first term, that is: $I = - \dfrac{\arcsin x × x}{\sqrt{1-x^2}}+ \ln (\sqrt{1-x^2}) + K $.
So, I want to know where that minus sign comes from.
Whilst waiting for your LaTeX improvement...
$$\int \frac{\arcsin(x)}{(\sqrt{1-x^2})^{3/2}}\ \text{d}x = $$
$$ = -\frac{\pi \left(1-x^2\right)^{3/4} \, _3F_2\left(\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};1-x^2\right)}{2 \sqrt{2} \Gamma \left(\frac{5}{4}\right) \Gamma \left(\frac{7}{4}\right)}-2 x \sqrt[4]{1-x^2} \, _2F_1\left(\frac{3}{4},1;\frac{5}{4};1-x^2\right) \sin ^{-1}(x)$$
Where $_aF_b$ are the hypergeometric functions, and $\Gamma(\cdot)$ is the Euler Gamma function.