$M = \int_0^{\pi/2} \frac{\cos x}{(x+2)} dx$ and $N = \int_0^{\pi/4} \frac{\sin x \cos x}{(x+1)^2} dx$, then value of $M - N$ is?

Question

Options are : (A) $\pi\quad$ (B) $\frac{\pi}{4}\quad$ (C) $\frac{2}{\pi + 4}\quad$ (D) $\frac{2}{\pi - 4}$

I have solved it using Taylor's expansion of the numerators. E.g, $\cos x = \cos ((x+2) -2) = \cos(x+2) \cos 2 + \sin(x+2) \sin 2$, then using Taylor's formula expanded $\cos(x+2)$ and $\sin(x+2)$ to get polynomials of $(x+2)$, which is in the denominator. But due to the denominator, I am getting results in terms of log, which does not match any of the options given.

Using NIntegrate in Mathematica didn't give anything close to the choices...are you sure you've stated the problem correctly? — Integrand, Jun 11 '20 at 05:22
@Integrand. $4$ is a typo. The upper bound must be $\frac \pi 4$. — Claude Leibovici, Jun 11 '20 at 07:50

Aditya Sriram · Accepted Answer · 2020-06-11T07:50:41.023

2

Using Integration by Parts, we can write N as $$N = -\frac{\sin x\cos x}{x+1}\vert_0^{\frac{\pi}{4}}+\int_0^\frac{\pi}{4}\frac{\cos^2x-\sin^2x}{x+1}dx=-\frac{2}{\pi+4}+\int_0^{\frac{\pi}{4}}\frac{\cos2x}{x+1}dx$$ In the second integral, let $x \to 2x \implies dx \to 2dx$ $$\therefore N=-\frac{2}{\pi+4}+\int_0^\frac{\pi}{2}\frac{\cos x}{x+2}dx=-\frac{2}{\pi+4}+M$$ $$\therefore M-N=\frac{2}{\pi+4}$$

P.S Thanks to Claude Leibovici for the corrected expression.

edited Jun 11 '20 at 07:50

answered Jun 11 '20 at 05:19

Aditya Sriram

1,292

Thanks Aditya for trying. Since $\frac{\pi}{2}$ < 8, the integral part might be a negative term. – Monalisha Bhowmik Jun 11 '20 at 07:12
Yeah, I agree with you. Like other people have already mentioned, there might be some error in the question :) – Aditya Sriram Jun 11 '20 at 07:13
I do not want to put an answer since you did most of the work. Do you mind to redo your calculation with an upper bound equal to $\frac \pi 4$ for $N$. Thanks and cheers. – Claude Leibovici Jun 11 '20 at 07:44
Sure! Will do so immediately! Thanks for finding the corrected expression :) – Aditya Sriram Jun 11 '20 at 07:46
You are very welcome ! Nice and simple and elegant solution. $\to +1$. – Claude Leibovici Jun 11 '20 at 07:52
Very nice solution. Thank you Aditya. – Monalisha Bhowmik Jun 12 '20 at 03:19
You're welcome :) – Aditya Sriram Jun 12 '20 at 03:51

score 2 · Answer 2 · answered Jun 11 '20 at 07:42

2

The real problem is $$N = \int_0^{\color{red}{\frac \pi 4}} \frac{\sin (x) \cos (x)}{(x+1)^2}\, dx$$ and the answer is $(C)$.

answered Jun 11 '20 at 07:42

Claude Leibovici

260,315

Thanks Claude for correcting. It might be a printing mistake in the paper. – Monalisha Bhowmik Jun 12 '20 at 03:18

$M = \int_0^{\pi/2} \frac{\cos x}{(x+2)} dx$ and $N = \int_0^{\pi/4} \frac{\sin x \cos x}{(x+1)^2} dx$, then value of $M - N$ is?

2 Answers2