Chapter 5.8, Problem 14E

To determine

To evaluate the integral ${\displaystyle \int \frac{\sin 2\theta }{\sqrt{5-\sin \theta }}d\theta }$

Answer to Problem 14E

The value of integral ${\displaystyle \int \frac{\sin 2\theta }{\sqrt{5-\sin \theta }}d\theta }$ is $2\left(\frac{1}{3}{\left(5-\sin \theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-5\sqrt{5-\sin \theta }\right)+C$

Explanation of Solution

Given information:

The given expression is ${\displaystyle \int \frac{\sin 2\theta }{\sqrt{5-\sin \theta }}d\theta }$

Formula used:

Sum rule

  ${\displaystyle \int f}\left(x\right)\pm g\left(x\right)dx={\displaystyle \int f}\left(x\right)dx\pm {\displaystyle \int g}\left(x\right)dx$

Calculation:

Consider, ${\displaystyle \int \frac{\sin 2\theta }{\sqrt{5-\sin \theta }}d\theta }$

  ${\displaystyle \int \frac{\sin 2\theta }{\sqrt{5-\sin \theta }}d\theta }={\displaystyle \int \frac{2\sin \theta \cos \theta }{\sqrt{5-\sin \theta }}d\theta }$

Apply u substitution: $u=\sin \theta \Rightarrow du=\cos \theta$

Also, $a=5,b=-1$

  ${\displaystyle \int \frac{2\sin \theta \cos \theta }{\sqrt{5-\sin \theta }}d\theta }=2{\displaystyle \int \frac{udu}{\sqrt{a+bu}}}$

Simplify, ${\displaystyle \int \frac{udu}{\sqrt{a+bu}}}$

Apply v substitution, $v=\sqrt{a+bu}$

  $={\displaystyle \int \frac{2\left(v^2-a\right)}{b^2}}dv$

Take the constant out: ${\displaystyle \int a}\cdot f\left(x\right)dx=a\cdot {\displaystyle \int f}\left(x\right)dx$

  $=\frac{2}{b^2}\cdot {\displaystyle \int }\left(v^2-a\right)dv$

Apply sum rule: ${\displaystyle \int f}\left(x\right)\pm g\left(x\right)dx={\displaystyle \int f}\left(x\right)dx\pm {\displaystyle \int g}\left(x\right)dx$

  $=\frac{2}{b^2}\left({\displaystyle \int }{v}^{2}dv-{\displaystyle \int }adv\right)$

  $=\frac{2}{b^2}\left(\frac{v^3}{3}-av\right)+C$ [Where C is constant]

Substitute back $v=\sqrt{a+bu}$

  $\begin{array}{l}=\frac{2}{b^2}\left(\frac{{\left(\sqrt{a+bu}\right)}^3}{3}-a\sqrt{a+bu}\right)+C\\ =\frac{2}{b^2}\left(\frac{1}{3}{\left(a+bu\right)}^{\frac{3}{2}}-a\sqrt{a+bu}\right)+C\end{array}$

Back substitute $u=\sin \theta ,a=5,b=-1$

  $\begin{array}{l}=\frac{2}{{\left(-1\right)}^2}\left(\frac{1}{3}{\left(5-\sin \theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-5\sqrt{5-\sin \theta }\right)+C\\ =2\left(\frac{1}{3}{\left(5-\sin \theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-5\sqrt{5-\sin \theta }\right)+C\end{array}$