int_0^{pi / 2} frac{sin x}{1+cos x+sin x} d x | vedclass

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

Question

(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi

Vedclass · Accepted Answer

$\frac{\pi}{4}-\frac{1}{2} \log 2$

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{1+\cos(\pi/2 - x)+\sin(\pi/2 - x)} dx = \int_0^{\pi / 2} \frac{\cos x}{1+\sin x+\cos x} dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{\sin x + \cos x}{1+\sin x+\cos x} dx$. $2I = \int_0^{\pi / 2} \frac{(1+\sin x+\cos x) - 1}{1+\sin x+\cos x} dx = \int_0^{\pi / 2} (1 - \frac{1}{1+\sin x+\cos x}) dx$. $2I = [x]_0^{\pi / 2} - \int_0^{\pi / 2} \frac{1}{1+2\sin(x/2)\cos(x/2)+2\cos^2(x/2)-1} dx$. $2I = \frac{\pi}{2} - \int_0^{\pi / 2} \frac{1}{2\cos^2(x/2)(	an(x/2)+1)} dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi / 2} \frac{\sec^2(x/2)}{	an(x/2)+1} dx$. Let $u = 	an(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$. $2I = \frac{\pi}{2} - \int_0^1 \frac{1}{u+1} du = \frac{\pi}{2} - [\log|u+1|]_0^1 = \frac{\pi}{2} - \log 2$. Therefore,$I = \frac{\pi}{4} - \frac{1}{2} \log 2$.

$\int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x=$

Similar Questions

If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx = \pi(\alpha \pi^2 + \beta)$,where $\alpha, \beta \in \mathbb{Z}$,then $(\alpha + \beta)^2$ equals:

$\left[ {\sum\limits_{n = 1}^{10} {\int_{ - 2n - 1}^{2n} {{{\sin }^{27}}x\,dx} } } \right] + \left[ {\sum\limits_{n = 1}^{10} {\int_{2n}^{2n + 1} {{{\sin }^{27}}x\,dx} } } \right]$ equals

Evaluate the integral: $\int_{-a}^{a} x^{2}\left(\frac{e^{x^{3}}-e^{-x^{3}}}{e^{x^{3}}+e^{-x^{3}}}\right) d x$

$\int_{-\pi / 2}^{\pi / 2}(2 \sin |x|+\cos |x|) d x=$

Consider $L = \sqrt[3]{2012} + \sqrt[3]{2013} + \ldots + \sqrt[3]{3011}$,$R = \sqrt[3]{2013} + \sqrt[3]{2014} + \ldots + \sqrt[3]{3012}$,and $I = \int_{2012}^{3012} \sqrt[3]{x} \, dx$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module