Suppose M = int_{0}^{pi / 2} frac{cos x}{x+2} | vedclass

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(D) Given,$M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Using the identity $\sin x \

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$\frac{2}{\pi+4}$

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(D) Given,$M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$,we have $N = \int_{0}^{\pi / 4} \frac{\sin 2x}{2(x+1)^{2}} dx$. Let $2x = t$,then $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\pi/4, t=\pi/2$. $N = \int_{0}^{\pi / 2} \frac{\sin t}{2(t/2 + 1)^{2}} \cdot \frac{dt}{2} = \int_{0}^{\pi / 2} \frac{\sin t}{2(\frac{t+2}{2})^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$. Now,$M - N = \int_{0}^{\pi / 2} \left( \frac{\cos x}{x+2} - \frac{\sin x}{(x+2)^{2}} \right) dx$. Using integration by parts on $\int \frac{\cos x}{x+2} dx$ with $u = \frac{1}{x+2}$ and $dv = \cos x dx$,we get $du = -\frac{1}{(x+2)^{2}} dx$ and $v = \sin x$. $M = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \left( -\frac{1}{(x+2)^{2}} \right) dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} + \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$. Thus,$M - \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} = \frac{\sin(\pi/2)}{\pi/2 + 2} - \frac{\sin(0)}{0+2} = \frac{1}{\frac{\pi+4}{2}} = \frac{2}{\pi+4}$.

Suppose $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$. Then,the value of $(M - N)$ equals

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