int_0^{pi / 2} frac{2 sin (x)+3 cos (x)}{sin | vedclass

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

Question

(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we ge

Vedclass · Accepted Answer

$\frac{5 \pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get: $I = \int_0^{\pi / 2} \frac{2 \sin (\pi/2-x)+3 \cos (\pi/2-x)}{\sin (\pi/2-x)+\cos (\pi/2-x)} d x = \int_0^{\pi / 2} \frac{2 \cos (x)+3 \sin (x)}{\cos (x)+\sin (x)} d x$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{(2 \sin x + 3 \cos x) + (2 \cos x + 3 \sin x)}{\sin x + \cos x} d x$. $2I = \int_0^{\pi / 2} \frac{5 \sin x + 5 \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 5 d x$. $2I = [5x]_0^{\pi / 2} = 5(\frac{\pi}{2} - 0) = \frac{5\pi}{2}$. Therefore,$I = \frac{5\pi}{4}$.

$\int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x=$

Similar Questions

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

$\int\limits_0^\infty {\frac{{{x^3}}}{{1 + x + 2{x^2} + 2{x^3} + {x^4} + {x^5}}}} dx$

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x=$

$\int_{0}^{100} e^{x-[x]} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module