Mathematics

# Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{4}}\dfrac{dx}{5+4 cos x}$

##### SOLUTION
$\displaystyle\int^{\pi/4}_0\dfrac{dx}{5+4\cos x}=\displaystyle\int^{\pi/4}_0\dfrac{dx}{5+4\left(\dfrac{1-\tan^2x/2}{1+\tan^2x/2}\right)}$
$=\displaystyle\int^{\pi/4}_0\dfrac{dx(1+\tan^2x/2)}{5+5\tan^2x/2+4-4\tan^2x/2}$
$=\displaystyle\int^{\pi/4}_0\dfrac{\sec^2x/2dx}{9+\tan^2x/2}$ Let $t=\tan x/2$                  $dt=\dfrac{1}{2}\sec^2x/2dx$
$=\displaystyle\int^{\pi/4}_0\dfrac{2dt}{9+t^2}$
When $x/2=0, t=0$; when $x/2=\pi/4$ $\Rightarrow t=1$
$=\displaystyle\int^1_0\dfrac{2dt}{9+t^2}$
$=\left.2\times \dfrac{1}{3}\tan^{-1}\left(\dfrac{t}{3}\right)\right]^1_0$
$=\dfrac{2}{3}\left[\tan^{-1}\left(1/3\right)-\tan^{-1}(0)\right]$
$=\dfrac{2}{3}\left[\tan^{-1}\left(1/3\right)-0\right]$
$=\dfrac{2}{3}\tan^{-1}\left(1/3\right)$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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