Question 4a(i)

SCH_CALC L3CALC

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$${\int }_0^1{\sin }^{-1}xdx\equiv {\int }_0^1{\sin }^{-1}x\cdot 1dx$$

Using integration by parts: $u={\sin }^{-1}x$ $u^{\prime }=?$ $v=x$ $v^{\prime }=1$

To get $u^{\prime }$:

$$y={\sin }^{-1}x⟺x=\sin y$$

Differentiate implicitly:

$$1=\cos y\cdot \frac{dy}{dx}⟺\frac{dy}{dx}=\frac{1}{\cos y}$$

Since $x=\sin y⟺x^2={\sin }^{2}y⟺-x^2=-1+{\cos }^{2}y$

$${\cos }^{2}y=1-x^2$$

We only use the positive root of $\cos y=\sqrt{1-x^2}$ because ${\sin }^{-1}x$ is defined within $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$, and because $\cos y$ between the interval $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$ is always positive (in the 1st and 4th quadrant)!

Hence

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\equiv u^{\prime }$$

Back to integration by parts:

$$\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$$ $$\int {\sin }^{-1}x\cdot 1dx=x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx$$

Calculate the integral on RHS ($\int \frac{x}{\sqrt{1-x^2}}dx$) using u-substitution:

$$=\int \frac{x}{\sqrt{u}}\cdot \frac{du}{-2x}$$ $$=\int \frac{1}{-2\sqrt{u}}du$$ $$=-\frac{1}{2}\int \frac{1}{\sqrt{u}}du$$

Back to integration by parts:

$$\int {\sin }^{-1}x\cdot 1dx=x{\sin }^{-1}x+\frac{1}{2}\int \frac{1}{\sqrt{u}}du$$ $$\int {\sin }^{-1}x\cdot 1dx=x{\sin }^{-1}x+\frac{1}{2}\int u^{-1/2}du$$ $$\int {\sin }^{-1}x\cdot 1dx=x{\sin }^{-1}x+\frac{1}{2}\left[2u^{\frac{1}{2}}\right]$$

Substitute $u=1-x^2$ back:

$$\int {\sin }^{-1}x\cdot 1dx=x{\sin }^{-1}x+\frac{1}{2}\left[2(1-x^2{)}^{\frac{1}{2}}\right]$$

Apply bounds from 0 to 1: NOTE:

$${\int }_a^{b}u{v}^{\prime }dx={\left[uv\right]}_a^b-{\int }_a^b{u}^{\prime }vdx$$ $${\int }_0^1{\sin }^{-1}x\cdot 1dx={\left[x{\sin }^{-1}x\right]}_0^1+\frac{1}{2}{\left[2(1-x^2{)}^{\frac{1}{2}}\right]}_0^1$$ $${\int }_0^1{\sin }^{-1}x\cdot 1dx=\boxed{\frac{\pi }{2}-1}$$

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Or alternatively what we can do is rotate the screen by 90 degrees, which will give us a $\sin x$ graph, and apply bounds from $-\frac{\pi }{2}$ and $\frac{\pi }{2}$!