The value of the integral int_{frac{pi}{4}}^{ | vedclass

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

Question

(B) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where

Vedclass · Accepted Answer

$\pi(\sqrt{2}-1)$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$: $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin(\pi-x)} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi-x}{1+\sin x} dx$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x + \pi - x}{1+\sin x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1+\sin x} dx$. Multiply numerator and denominator by $(1-\sin x)$: $2I = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1-\sin x}{\cos^2 x} dx = \pi \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (\sec^2 x - \sec x 	an x) dx$. Integrating: $2I = \pi [	an x - \sec x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$. Evaluating at the limits: $2I = \pi [(	an \frac{3\pi}{4} - \sec \frac{3\pi}{4}) - (	an \frac{\pi}{4} - \sec \frac{\pi}{4})]$. $2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2]$. $I = \pi(\sqrt{2}-1)$.

The value of the integral $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x}{1+\sin x} dx$ is

Similar Questions

Show that $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,if $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

$\int_{-4}^{4} (2^x + 2^{-x})(3^x + 3^{-x}) \, dx$ is equal to

The value of $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is

If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 2$,then $\int_0^a f(x)g(x) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module