Integration by Parts — Question 7

Question 7

Find an antiderivative $F$ of $$f(x)=\arcsin(x), \qquad -1<x<1,$$ that satisfies the initial condition $$F(0)=2.$$

Question 7 – Solution

We begin by evaluating $$\int \arcsin(x)\,dx.$$ Since the integrand can be viewed as $\arcsin(x)\cdot1$, use integration by parts with $$u=\arcsin(x),\quad dv=dx, \qquad du=\frac{1}{\sqrt{1-x^2}}\,dx,\quad v=x.$$ Using $\int u\,dv=uv-\int v\,du$, we obtain $$\begin{align*} \int \arcsin(x)\,dx &=x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}\,dx. \tag{1} \end{align*}$$

To evaluate the remaining integral, let $$w=1-x^2.$$ Then $$dw=-2x\,dx, \qquad x\,dx=-\frac12\,dw.$$ Therefore, $$\begin{align*} \int\frac{x}{\sqrt{1-x^2}}\,dx &=-\frac12\int w^{-1/2}\,dw \\ &=-\frac12\left(2w^{1/2}\right) \\ &=-\sqrt{w} \\ &=-\sqrt{1-x^2}. \end{align*}$$ Substitute this result into equation (1): $$\begin{align*} \int \arcsin(x)\,dx &=x\arcsin(x)-\left(-\sqrt{1-x^2}\right)+C \\ &=x\arcsin(x)+\sqrt{1-x^2}+C. \end{align*}$$

Thus, the general antiderivative is $$F(x)=x\arcsin(x)+\sqrt{1-x^2}+C.$$ Now apply the condition $F(0)=2$: $$\begin{align*} F(0) &=0\cdot\arcsin(0)+\sqrt{1-0^2}+C \\ &=1+C. \end{align*}$$ Hence, $$1+C=2 \qquad\Longrightarrow\qquad C=1.$$ Therefore, the required antiderivative is $$\boxed{\displaystyle F(x)=x\arcsin(x)+\sqrt{1-x^2}+1}.$$

To verify, differentiate: $$\begin{align*} F'(x) &=\arcsin(x)+\frac{x}{\sqrt{1-x^2}} -\frac{x}{\sqrt{1-x^2}} \\ &=\arcsin(x). \end{align*}$$ Also, $F(0)=0+1+1=2$, so both requirements are satisfied.