The integral int_0^pi frac{(x+3) sin x}{1+3 c | vedclass

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Question

(C) Let $I = \int_0^\pi \frac{(x+3) \sin x}{1+3 \cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi \fra

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$\frac{\pi}{3 \sqrt{3}}(\pi+6)$

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(C) Let $I = \int_0^\pi \frac{(x+3) \sin x}{1+3 \cos^2 x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^\pi \frac{(\pi-x+3) \sin(\pi-x)}{1+3 \cos^2(\pi-x)} dx = \int_0^\pi \frac{(\pi-x+3) \sin x}{1+3 \cos^2 x} dx$. Adding the two expressions for $I$: $2I = \int_0^\pi \frac{(x+3+\pi-x+3) \sin x}{1+3 \cos^2 x} dx = \int_0^\pi \frac{(\pi+6) \sin x}{1+3 \cos^2 x} dx$. Since $\sin(\pi-x) = \sin x$ and $\cos^2(\pi-x) = \cos^2 x$,we use the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$: $2I = 2(\pi+6) \int_0^{\pi/2} \frac{\sin x}{1+3 \cos^2 x} dx$. $I = (\pi+6) \int_0^{\pi/2} \frac{\sin x}{1+3 \cos^2 x} dx$. Let $\sqrt{3} \cos x = t$,then $-\sqrt{3} \sin x dx = dt$,so $\sin x dx = -\frac{dt}{\sqrt{3}}$. When $x=0, t=\sqrt{3}$. When $x=\pi/2, t=0$. $I = (\pi+6) \int_{\sqrt{3}}^0 \frac{-dt/\sqrt{3}}{1+t^2} = \frac{\pi+6}{\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1+t^2}$. $I = \frac{\pi+6}{\sqrt{3}} [	an^{-1} t]_0^{\sqrt{3}} = \frac{\pi+6}{\sqrt{3}} (	an^{-1} \sqrt{3} - 	an^{-1} 0) = \frac{\pi+6}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi(\pi+6)}{3\sqrt{3}}$.

The integral $\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x$ is equal to :

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