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
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