int_0^{frac{pi}{4}} log (1+tan x) , dx = | vedclass

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

Question

(D) ધારો કે $I = \int_0^{\frac{\pi}{4}} \log (1+	an x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{\fra

Vedclass · Accepted Answer

$\frac{\pi}{8} \log 2$

Vedclass · Answer

(D) ધારો કે $I = \int_0^{\frac{\pi}{4}} \log (1+	an x) \, dx$. ગુણધર્મ $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ નો ઉપયોગ કરતા: $I = \int_0^{\frac{\pi}{4}} \log \left(1+	an \left(\frac{\pi}{4}-xight)ight) \, dx$. કારણ કે $	an(\frac{\pi}{4}-x) = \frac{1-	an x}{1+	an x}$,તેથી: $I = \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-	an x}{1+	an x}ight) \, dx = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+	an x}ight) \, dx$. $I = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1+	an x)) \, dx$. $I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1+	an x) \, dx$. $I = [x \log 2]_0^{\frac{\pi}{4}} - I$. $2I = \frac{\pi}{4} \log 2$. $I = \frac{\pi}{8} \log 2$.

$\int_0^{\frac{\pi}{4}} \log (1+\tan x) \, dx =$

Similar Questions

સંકલન $\int_{-\pi / 2}^{\pi / 2}\left(x^2+\ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$ નું મૂલ્ય શોધો.

જો $I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ હોય,તો $\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x$ ની કિંમત શોધો:

$\int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$ ની કિંમત શોધો.

નિશ્ચિત સંકલન $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module