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Question

Let $$g(x)=\displaystyle\dfrac{1+2\cos x}{(\cos x+2)^2}dx$$ and $$g(0)=0$$ then value of $$32 g\left(\dfrac{\pi}{2}\right)$$ is?


Solution

$$g(x) = \displaystyle \int \cfrac{1+2\cos x}{(\cos x+2)^2} dx $$
Let $$t = \tan \cfrac{x}{2}$$
$$dt = \cfrac{1}{2} \sec^2 \cfrac{x}{2} dx = \cfrac{1}{2} (t^2+1)dx$$
$$\sin x = \cfrac{2t}{1+t^2} , \cos x = \cfrac{1-t^2}{1+t^2}$$
$$\displaystyle 2\int \cfrac{3-t^2}{(t^2+3)^2} dt$$
$$\implies \cfrac{2t}{t^2+3} + C = \cfrac{2\tan \cfrac{x}{2}}{\tan^2\cfrac{x}{2} + 3} + C$$
At x = 0, we have g(0) = 0,or
$$C = 0$$
At $$x = \cfrac{\pi}{2}$$, we have
$$\cfrac{2}{1+3} . 32 = 16$$ 

Mathematics

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