Chapter 5.7, Problem 36E

To determine

To calculate: The value of given integral by completing the square in the quadratic $3-2x-x^2$ and making a substitution

Answer to Problem 36E

The value of the given integral is $-\sqrt{\left(1-x\right)\left(3-x\right)}-{\sin }^{-1}\frac{x+1}{2}+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{x}{\sqrt{3-2x-x^2}}dx}$

Formula used:

  ${\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}x+C}$

Calculation:

The given integral is

  ${\displaystyle \int \frac{x}{\sqrt{3-2x-x^2}}dx}$

At first complete the square in the denominator, so

  $\begin{array}{l}{\displaystyle \int \frac{x}{\sqrt{3-2x-x^2}}dx}\\ ={\displaystyle \int \frac{x}{\sqrt{-\left(x^2+2x+1-4\right)}}dx}\\ ={\displaystyle \int \frac{x}{\sqrt{2^2-{\left(x+1\right)}^2}}dx}\end{array}$

Use the substitution $x+1=2\sin \theta$

  $\begin{array}{l}x+1=2\sin \theta \\ \Rightarrow {\left(x+1\right)}^2=4{\sin }^2\theta \\ \Rightarrow x=2\sin \theta -1\\ \Rightarrow \theta ={\sin }^{-1}\frac{x+1}{2}\\ \Rightarrow d\left(x+1\right)=d\left(2\sin \theta \right)\left[\text{differentiate both sides}\right]\\ \Rightarrow d\left(x\right)=2\cos \theta d\theta \end{array}$

Now, the integral becomes,

  ${\displaystyle \int \frac{2\sin \theta -1}{\sqrt{4{\sin }^2\theta -4}}2\cos \theta d\theta }$

Use the identities ${\cos }^{2}x+{\sin }^{2}x=1$ to solve the integral

  $\begin{array}{l}{\displaystyle \int \frac{2\sin \theta -1}{\sqrt{4{\sin }^2\theta -4}}2\cos \theta d\theta }\\ ={\displaystyle \int \frac{2\sin \theta -1}{\cos \theta }}2\cos \theta d\theta \\ ={\displaystyle \int \left(4\sin \theta -2d\theta \right)}\\ =-4\cos \theta -2\theta +C\end{array}$

Now put the value $\theta ={\sin }^{-1}\frac{x+1}{2}$ and calculate the final result

  $\begin{array}{l}{\displaystyle \int \frac{2\sin \theta -1}{\sqrt{4{\sin }^2\theta -4}}2\cos \theta d\theta }\\ ={\displaystyle \int \frac{2\sin \theta -1}{2\cos \theta }}2\cos \theta d\theta \\ ={\displaystyle \int \left(2\sin \theta -1\right)}d\theta \\ =-2\cos \left({\sin }^{-1}\frac{x+1}{2}\right)-{\sin }^{-1}\frac{x+1}{2}+C\\ =-\sqrt{4-{\left(x+1\right)}^2}-{\sin }^{-1}\frac{x+1}{2}+C\\ =-\sqrt{\left(2-x-1\right)\left(2-x+1\right)}-{\sin }^{-1}\frac{x+1}{2}+C\\ =-\sqrt{\left(1-x\right)\left(3-x\right)}-{\sin }^{-1}\frac{x+1}{2}+C\end{array}$

Therefore, the value of the given integral is $-\sqrt{\left(1-x\right)\left(3-x\right)}-{\sin }^{-1}\frac{x+1}{2}+C$