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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) dx = (\int_0^1 f(x) dx)^2$. Then,the range of $f$

$\int_{\,\pi }^{\,10\pi } {|\sin x|dx}$ is

Let $\int_{-2}^{2} (\sin |x| + [x \sin x]) dx = 2(3 - \cos 2) + \beta$, where $[\cdot]$ denotes the greatest integer function. Then $\beta \sin \left(\frac{\beta}{2}\right)$ equals:

If $f$ and $g$ are continuous functions in $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 4$,then $\int_{0}^{a} f(x) g(x) dx$ is equal to:

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module