The value of int_0^{pi /2} {log left( {frac{{ | vedclass

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

Question

(C) Let $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin(\pi/2 - x)}}{{4 + 3\cos(\pi/2 - x)}}} \right)} \,dx$. Since $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$,we have: $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \,dx$. Using the property $\log(a/b) = -\log(b/a)$,we get: $I = \int_0^{\pi /2} -\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \,dx = -I$. Therefore,$2I = 0$,which implies $I = 0$.

The value of $\int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$ is

Similar Questions

$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to

The value of $\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ is :

The value of $\int_{ - 1}^1 {\frac{{\sin x - {x^2}}}{{3 - |x|}}\,dx} $ is

$\int_0^2 x^2(2-x)^5 d x=$

$\int_{-1}^3 \left(\tan^{-1}\left(\frac{x}{x^2+1}\right) + \tan^{-1}\left(\frac{x^2+1}{x}\right)\right) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module