Question

Evaluate the following definite integral:

${\int }_0^{{\pi }_{/}2}\frac{1}{4+5\cos x}dx=$

• A. $\frac{1}{5}\log 2$
• B. $\frac{1}{2}\log 2$
• C. $\frac{1}{3}\log 3$
• D. $\frac{1}{3}\log 2$

Answer 1

Answer: (D)

$\frac{1}{3}\log 2$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{1}{4+5cosx}dx$

Substitute $tan\left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{sec}^2\left(\frac{x}{2}\right)dx=dt$

So, $x\to 0\Rightarrow t\to 0$ and $x\to \frac{\pi }{2}\Rightarrow t\to 1$

gives $\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$

$\Rightarrow$ $I={\int }_0^1\frac{1}{4+5\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=2{\int }_0^1\frac{1}{9-t^2}dt$

$\Rightarrow$ $I=\frac{2}{3}{\int }_0^1\frac{1}{1-\frac{t^2}{9}}dt$

Substitute $\frac{t}{3}=u\Rightarrow dt=3du$

So, $t\to 0\Rightarrow u\to 0$ and $t\to 1\Rightarrow u\to \frac{1}{3}$

$\Rightarrow$ $I=\frac{2}{3}{\int }_0^{\frac{1}{3}}\frac{1}{1-u^2}du=\frac{2}{3}{\left[\frac{1}{2}\log \frac{1+u}{1-u}\right]}_0^{\frac{1}{3}}$

$\Rightarrow$ $I=\frac{1}{3}(\log 2-0)=\frac{1}{3}\log 2$