Inverse Trig Integral

we make substitution of $\sin^{-1} x =u$ which gives us $\,dx = \cos u\,du$ . Thus we have $I=\int_{0}^{\frac{1}{2}\pi}u\cos u \cos^{-1} \left(\cos \left(\frac{\pi}{2}-u\right)\right)\,du =\dfrac{\pi}{2}\int_{0}^{\frac{1}{2}\pi}u \cos u \,du-\int_{0}^{\frac{1}{2}\pi} u^2\cos u\,du$ Using IBP we evalute the integrals as $\dfrac{\pi}{2}\int_{0}^{\frac{1}{2}\pi}u \cos u \,du =\dfrac{\pi^2}{4}-\dfrac{\pi}{2}$ and using IBP twice $\int_{0}^{\frac{1}{2}\pi} u^2\cos u\,du = \dfrac{\pi^2}{4} -2$ which gives us $I= \left(\dfrac{\pi^2}{4} -\dfrac{\pi}{2}\right) -\left(\dfrac{\pi^2}{4} -2 \right)=2-\dfrac{\pi}{2}$ thus $\dfrac{a}{b} =-4$

Here is the generalized problem.

$\begin{aligned} I & = \int_0^1 \sin^{-1} x \cos^{-1} x \ dx & \small \color{#3D99F6} \text{Note that } \cos^{-1} x = \frac \pi 2 - \sin^{-1} x \\ & = \frac \pi 2 \int_0^1 \sin^{-1} x \ dx - \int_0^1 \left(\sin^{-1} x\right)^2 dx & \small \color{#3D99F6} \text{By integration by parts} \end{aligned}$

$\begin{aligned} \ \ & = \frac {\pi x \sin^{-1} x}2 \bigg|_0^1 - \int_0^1 \frac {\pi x}{2\sqrt{1-x^2}} dx - x (\sin^{-1} x)^2 \bigg|_0^1 + \int_0^1 \frac {2x\sin^{-1}x}{\sqrt{1-x^2}} dx \\ & = \frac {\pi^2}4 + \frac {\pi \sqrt{1-x^2}}2 \bigg|_0^1 - \frac {\pi^2}4 - 2 \sqrt{1-x^2} \sin^{-1} x\bigg|_0^1 + 2 \int_0^1 \ dx \\ & = - \frac \pi 2 - 0 + 2 = 2 - \frac \pi 2 \end{aligned}$

Therefore, $\dfrac ab = \dfrac 2{-\frac 12} = \boxed{-4}$ .

We have, using the substitution $y = \cos2\theta$ , $\begin{aligned} \int_0^1 \sin^{-1}x\,dx & = \; \int_0^1 \int_0^x \frac{dy}{\sqrt{1-y^2}}\,dx \; = \; \int_0^1 \int_y^1\,dx \frac{dy}{\sqrt{1-y^2}} \; = \; \int_0^1 \frac{1-y}{\sqrt{1-y^2}}\,dy \; = \; \int_0^1 \sqrt{\frac{1-y}{1+y}}\,dy \\ & = \; \int_0^{\frac14\pi} \tan\theta \times 2\sin2\theta\,d\theta \;= \; 4\int_0^{\frac14\pi}\sin^2\theta\,d\theta \; = \; \Big[2\theta - \sin2\theta\Big]_0^{\frac14\pi} \; = \; \tfrac12\pi - 1 \end{aligned}$ and $\begin{aligned} \int_0^1 (\sin^{-1}x)^2\,dx & = \; \Big[x(\sin^{-1}x)^2\Big]_0^1 - 2\int_0^1 \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx \; = \; \tfrac14\pi^2 - 2\int_0^1 \frac{x}{\sqrt{1-x^2}}\int_0^x \frac{dy}{\sqrt{1-y^2}}\,dx \\ & = \; \tfrac14\pi^2 - 2\int_0^1 \frac{1}{\sqrt{1-y^2}}\int_y^1 \frac{x\,dx}{\sqrt{1-x^2}}\,dy \; = \; \tfrac14\pi^2 - 2\int_0^1 \frac{1}{\sqrt{1-y^2}}\Big[-\sqrt{1-x^2}\Big]_y^1\,dy \\ & = \; \tfrac14\pi^2 - 2\int_0^1\,dy \; = \; \tfrac14\pi^2 - 2 \end{aligned}$ so that $\int_0^1 \sin^{-1}x\,\cos^{-1}x\,dx \; = \; \int_0^1 \sin^{-1}x\left(\tfrac12\pi - \sin^{-1}x\right)\,dx \; = \; \tfrac12\pi\left(\tfrac12\pi - 1\right) - \left(\tfrac14\pi^2 - 2\right) \; = \; 2 - \tfrac12\pi$ making $a = 2$ , $b = -\tfrac12$ , and hence $\tfrac{a}{b} = \boxed{-4}$ .