The value of the integral int_{frac{pi}{6}}^{ | vedclass

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

Question

(A) Let $I = \int_{\pi/6}^{\pi/3} \frac{\sin x - x \cos x}{x(x + \sin x)} dx$We can rewrite the numerator as $(x + \sin x) - x(1 + \cos x)$.So, $I = \

Vedclass · Accepted Answer

$\log_{e} \left\{ \frac{2(\pi + 3)}{(2\pi + 3\sqrt{3})} \right\}$

Vedclass · Answer

(A) Let $I = \int_{\pi/6}^{\pi/3} \frac{\sin x - x \cos x}{x(x + \sin x)} dx$We can rewrite the numerator as $(x + \sin x) - x(1 + \cos x)$.So, $I = \int_{\pi/6}^{\pi/3} \left( \frac{x + \sin x}{x(x + \sin x)} - \frac{x(1 + \cos x)}{x(x + \sin x)} \right) dx$$I = \int_{\pi/6}^{\pi/3} \left( \frac{1}{x} - \frac{1 + \cos x}{x + \sin x} \right) dx$Integrating term by term: $I = [\log |x|]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} \frac{1 + \cos x}{x + \sin x} dx$For the second integral, let $t = x + \sin x$, then $dt = (1 + \cos x) dx$.When $x = \pi/6$, $t = \pi/6 + 1/2 = (\pi + 3)/6$. When $x = \pi/3$, $t = \pi/3 + \sqrt{3}/2 = (2\pi + 3\sqrt{3})/6$.$I = \log(\pi/3) - \log(\pi/6) - [\log |t|]_{(\pi+3)/6}^{(2\pi+3\sqrt{3})/6}$$I = \log(2) - \left( \log \left( \frac{2\pi + 3\sqrt{3}}{6} \right) - \log \left( \frac{\pi + 3}{6} \right) \right)$$I = \log(2) - \log \left( \frac{2\pi + 3\sqrt{3}}{\pi + 3} \right) = \log \left( \frac{2(\pi + 3)}{2\pi + 3\sqrt{3}} \right)$

The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x - x \cos x)}{x(x + \sin x)} dx$ is

Similar Questions

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

Evaluate the definite integral $\int_0^4 x[x] \, dx$,where $[x]$ denotes the greatest integer function not greater than $x$.

Evaluate the definite integral $\int_{-2}^{2.24} [x] \, dx$,where $[x]$ denotes the greatest integer function.

If $f(x) = \operatorname{Max}\{x^3-4, x^4-4\}$ and $g(x) = \operatorname{Min}\{x^2, x^3\}$,then $\int_{-1}^1 (f(x) - g(x)) \, dx =$

Let $f(x) = |x - 2|$ and $g(x) = f(f(x))$,$x \in [0, 4]$. Then $\int_{0}^{3} (g(x) - f(x)) \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module