If int_{0}^{pi} log (sin x) dx = 8 k,then int | vedclass

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

Question

(B) Let $I = \int_{0}^{\pi / 4} \log (1 + 	an x) dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:$I = \int_{0}^{\pi

Vedclass · Accepted Answer

$-k$

Vedclass · Answer

(B) Let $I = \int_{0}^{\pi / 4} \log (1 + 	an x) dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:$I = \int_{0}^{\pi / 4} \log (1 + 	an (\frac{\pi}{4} - x)) dx$.Since $	an (\frac{\pi}{4} - x) = \frac{1 - 	an x}{1 + 	an x}$,we get:$I = \int_{0}^{\pi / 4} \log (1 + \frac{1 - 	an x}{1 + 	an x}) dx = \int_{0}^{\pi / 4} \log (\frac{2}{1 + 	an x}) dx$.$I = \int_{0}^{\pi / 4} (\log 2 - \log (1 + 	an x)) dx = \frac{\pi}{4} \log 2 - I$.$2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2$.Given $\int_{0}^{\pi} \log (\sin x) dx = 8k$. We know $\int_{0}^{\pi} \log (\sin x) dx = -\pi \log 2$.So,$8k = -\pi \log 2 \implies \pi \log 2 = -8k$.Substituting this into $I = \frac{\pi}{8} \log 2$,we get $I = \frac{1}{8} (-8k) = -k$.

If $\int_{0}^{\pi} \log (\sin x) dx = 8 k$,then $\int_{0}^{\pi / 4} \log (1 + \tan x) dx =$

Similar Questions

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

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$.

The value of $\int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$ is equal to: (in $\pi$)

If $f(a + b - x) = f(x)$,then $\int_a^b x f(x) dx = $

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module