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If $M=\int_{0}^{n / 2} \frac{\cos x}{x+2} d x, N=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x,$ then
the value of $M-N$ is
Options:
the value of $M-N$ is
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Verified Answer
The correct answer is:
$\frac{2}{\pi+4}$
Given, $M=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(x+2)} d x$
$\begin{aligned} \text { and } \quad N &=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x \\ &=\int_{0}^{\pi / 4} \frac{1}{2} \cdot \frac{\sin 2 x}{(x+1)^{2}} d x \end{aligned}$
Put $2 x=t \Rightarrow d x=\frac{a}{2}$
$\begin{aligned} \therefore \quad N &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{4(t / 2+1)^{2}} d t \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(t+2)^{2}} d t \end{aligned}$
$\therefore M-N=\int_{0}^{\frac{\pi}{2}}(\cos x) \cdot \frac{1}{(x+2)} d x$
$-\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} d x$
$=\left(\frac{\sin x}{x+2}\right)_{0}^{\pi 2}-\int_{0}^{\pi 2}-\frac{\sin x}{(x+2)^{2}} d x$
$-\frac{\sin \pi / 2}{\pi / 2+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$
$\begin{aligned} \text { and } \quad N &=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x \\ &=\int_{0}^{\pi / 4} \frac{1}{2} \cdot \frac{\sin 2 x}{(x+1)^{2}} d x \end{aligned}$
Put $2 x=t \Rightarrow d x=\frac{a}{2}$
$\begin{aligned} \therefore \quad N &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{4(t / 2+1)^{2}} d t \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(t+2)^{2}} d t \end{aligned}$
$\therefore M-N=\int_{0}^{\frac{\pi}{2}}(\cos x) \cdot \frac{1}{(x+2)} d x$
$-\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} d x$
$=\left(\frac{\sin x}{x+2}\right)_{0}^{\pi 2}-\int_{0}^{\pi 2}-\frac{\sin x}{(x+2)^{2}} d x$
$-\frac{\sin \pi / 2}{\pi / 2+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$
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