Question

Evaluate: $\underset{0}{\overset{\pi /2}{\int }}\frac{1}{2cosx+4sinx}dx$

• A. $\sqrt{5}log\left(\frac{3+\sqrt{5}}{2}\right)$
• B. $\frac{1}{\sqrt{5}}log\left(\frac{3-\sqrt{5}}{2}\right)$
• C. $\frac{1}{\sqrt{5}}log\left(\frac{3+\sqrt{5}}{2}\right)$
• D. None of these

Answer 1

Answer: (C)

We have, $I=\underset{0}{\overset{\pi /2}{\int }}\frac{1}{2cosx+4sinx}dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{1}{2\left(\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)+4\left(\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)}dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{se{c}^2\frac{x}{2}}{2-2ta{n}^2\frac{x}{2}+8tan\frac{x}{2}}dx$

Let tan $\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt$

Also, $x=0\Rightarrow t=0andx=\frac{\pi }{2}\Rightarrow t=1$

$\therefore I=\underset{0}{\overset{1}{\int }}\frac{2dt}{2-2t^2+8t}$

$=\underset{0}{\overset{1}{\int }}\frac{dt}{-\left[{\left(t-2\right)}^2-5\right]}=\underset{0}{\overset{1}{\int }}\frac{dt}{{\left(\sqrt{5}\right)}^2-{\left(t-2\right)}^2}dt$

$\Rightarrow I=\frac{1}{2\sqrt{5}}{\left[log\left|\sqrt{5}+t-2\right|\right]}_0^1$

$=\frac{1}{2\sqrt{5}}\left[log\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)-log\left(\frac{\sqrt{5}-2}{\sqrt{5}+2}\right)\right]$

$=\frac{1}{2\sqrt{5}}log\left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\right)=\frac{1}{2\sqrt{5}}log{\left(\frac{3+\sqrt{5}}{2}\right)}^2$

$\Rightarrow I=\frac{1}{\sqrt{5}}log\left(\frac{3+\sqrt{5}}{2}\right)$