Hello, i am struggling with this integration, I know how to do either sin²(x)dx or arcsin(x)dx, but not both at the same time.
Could someone point me in the right direction? I would really appreciate it
regards,
lazarus
As that substitution is not entirely obvious, as an alternative, it is possible to do this integral directly with integration by parts. Let \(\displaystyle \displaystyle \begin{align*} u = \arcsin^2{(x)} \implies \mathrm{d}u = \frac{2\arcsin{(x)}}{\sqrt{1 - x^2}}\,\mathrm{d}x \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \mathrm{d}v = x\,\mathrm{d}x \implies v = \frac{x^2}{2} \end{align*}\) and the integral becomes
\(\displaystyle \displaystyle \begin{align*} \int{x\arcsin^2{(x)}\,\mathrm{d}x} &= \frac{x^2\arcsin^2{(x)}}{2} - \int{ \frac{2\,x^2\arcsin{(x)}}{\sqrt{1 - x^2}}\,\mathrm{d}x } \end{align*}\)
Now apply integration by parts again with \(\displaystyle \displaystyle \begin{align*} u = \arcsin{(x)} \implies \mathrm{d}u = \frac{1}{\sqrt{1 - x^2}}\,\mathrm{d}x \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \mathrm{d}v = \frac{2\,x^2}{\sqrt{1 - x^2}}\,\mathrm{d}x \end{align*}\). Finding \(\displaystyle \displaystyle \begin{align*} v \end{align*}\) is tough, but not impossible.
\(\displaystyle \displaystyle \begin{align*} v &= \int{ \frac{2\,x^2}{\sqrt{1 - x^2}}\,\mathrm{d}x } \\ &= - \left( 2\,x\,\sqrt{1 - x^2} - \int{ 2\,\sqrt{1 - x^2}\,\mathrm{d}x } \right) \\ &= -2\,x\,\sqrt{1 - x^2} + 2\int{ \sqrt{1 - x^2}\,\mathrm{d}x } \\ &= -2\,x\,\sqrt{1 - x^2} + 2\int{ \sqrt{1 - \sin^2{(\theta )}}\,\cos{(\theta )} \,\mathrm{d}\theta } \textrm{ after substituting } x = \sin{(\theta)} \implies \mathrm{d}x = \cos{(\theta)}\,\mathrm{d}\theta \\ &= -2\,x\,\sqrt{1 - x^2} + 2\int{ \cos^2{(\theta )} \,\mathrm{d}\theta } \\ &= -2\,x\,\sqrt{1 - x^2} + 2\int{ \frac{1}{2}\,\left[ 1 + \cos{( 2\,\theta ) } \right] \,\mathrm{d}\theta } \\ &= -2\,x\,\sqrt{1 - x^2} + \int{ \left[ 1 + \cos{ \left( 2\,\theta \right) } \right] \,\mathrm{d}\theta } \\ &= -2\,x\,\sqrt{1 - x^2} + \theta +\frac{1}{2}\sin{ \left( 2\,\theta \right) } \\ &= -2\,x\,\sqrt{1 - x^2} + \arcsin{(x)} + \sin{\left( \theta \right) } \cos{ \left( \theta \right) } \\ &= -2\,x\,\sqrt{1 - x^2} + \arcsin{(x)} + \sin{ \left( \theta \right) } \,\sqrt{ 1 - \sin^2{ \left( \theta \right) } } \\ &= -2\,x\,\sqrt{1 - x^2} + \arcsin{(x)} + x\,\sqrt{ 1 - x^2 } \\ &= \arcsin{(x)} - x\,\sqrt{1 - x^2} \end{align*}\)
and thus going back to the original integral
\(\displaystyle \displaystyle \begin{align*} \frac{x^2\arcsin^2{(x)}}{2} - \int{ \frac{2\,x^2\arcsin{(x)}}{\sqrt{1 - x^2}}\,\mathrm{d}x } &= \frac{x^2\arcsin^2{(x)}}{2} - \left\{ \arcsin{(x)}\,\left[ \arcsin{(x)} - x\,\sqrt{1 - x^2} \right] - \int{ \left[ \arcsin{(x)} - x\,\sqrt{1 - x^2} \right]\,\frac{1}{\sqrt{1 - x^2}} \,\mathrm{d}x } \right\} \\ &= \frac{x^2\arcsin^2{(x)}}{2} - \arcsin^2{(x)} + x\,\sqrt{1 - x^2}\arcsin{(x)} + \int{ \left[ \frac{\arcsin{(x)}}{\sqrt{1 - x^2}} - x \right] \,\mathrm{d}x } \\ &= \frac{x^2\arcsin^2{(x)}}{2} - \arcsin^2{(x)} + x\,\sqrt{1 - x^2}\arcsin{(x)} + \frac{\arcsin^2{(x)}}{2} - \frac{x^2}{2} + C \\ &= \frac{x^2\arcsin^2{(x)} - \arcsin^2{(x)} + 2\,x\,\sqrt{1 - x^2}\arcsin{(x)} - x^2}{2} + C \end{align*}\)
But the approach by Subhotosh Khan is much more elegant. Why not try that method and see if you get the same answer (and in fewer steps with fewer difficult integrations)?
