int_0^{frac{pi}{4}} frac{sec ^2 x}{(1+tan x)( | vedclass

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

Question

(C) Let $1+	an x = t$. Then $\sec^2 x \, dx = dt$. When $x = 0$,$t = 1+	an(0) = 1$. When $x = \frac{\pi}{4}$,$t = 1+	an(\frac{\pi}{4}) = 2$. The in

Vedclass · Accepted Answer

$\log \left(\frac{4}{3}\right)$

Vedclass · Answer

(C) Let $1+	an x = t$. Then $\sec^2 x \, dx = dt$. When $x = 0$,$t = 1+	an(0) = 1$. When $x = \frac{\pi}{4}$,$t = 1+	an(\frac{\pi}{4}) = 2$. The integral becomes: $\int_1^2 \frac{dt}{t(t+1)}$. Using partial fractions: $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$. So,the integral is $\int_1^2 \left( \frac{1}{t} - \frac{1}{t+1} ight) dt$. $= [\log|t| - \log|t+1|]_1^2 = [\log|\frac{t}{t+1}|]_1^2$. $= \log(\frac{2}{3}) - \log(\frac{1}{2}) = \log(\frac{2/3}{1/2}) = \log(\frac{4}{3})$.

$\int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{(1+\tan x)(2+\tan x)} d x=$

Similar Questions

If $\int_{\log 2}^x \frac{du}{({e^u} - 1)^{1/2}} = \frac{\pi}{6}$,then ${e^x} = $

$\int_0^{\pi /4} \tan^6 x \sec^2 x \, dx = $

Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$.

The value of $\int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$ is

Evaluate the following integral: $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$

Vedclass Test Series

Exam Paper Generator

Online Exam Module