int_0^{frac{pi}{4}} x sec^2 x , dx = | vedclass

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

Question

(B) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x$ and $dv = \sec^2 x \, dx$. Then $du = dx$ and $v = 	an

Vedclass · Accepted Answer

$\frac{\pi}{4} - \log \sqrt{2}$

Vedclass · Answer

(B) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x$ and $dv = \sec^2 x \, dx$. Then $du = dx$ and $v = 	an x$. Applying the formula: $\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = [x 	an x]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} 	an x \, dx$. Evaluating the first term: $[x 	an x]_0^{\frac{\pi}{4}} = (\frac{\pi}{4} 	an \frac{\pi}{4}) - (0 \cdot 	an 0) = \frac{\pi}{4} \cdot 1 - 0 = \frac{\pi}{4}$. Evaluating the integral of $	an x$: $\int 	an x \, dx = \ln |\sec x|$. So,$\int_0^{\frac{\pi}{4}} 	an x \, dx = [\ln |\sec x|]_0^{\frac{\pi}{4}} = \ln |\sec \frac{\pi}{4}| - \ln |\sec 0| = \ln \sqrt{2} - \ln 1 = \ln \sqrt{2} - 0 = \ln \sqrt{2}$. Combining the results: $\frac{\pi}{4} - \ln \sqrt{2}$.

$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx =$

Similar Questions

Domain of definition of the function $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$ is

The function $L(x) = \int_1^x \frac{dt}{t}$ satisfies the equation

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

$\int_0^{\pi / 2} \frac{\pi \sin x}{1+\cos ^2 x} d x$ is equal to

If $f(x) = e^{-x} + 2e^{-2x} + 3e^{-3x} + \dots + \infty$,then $\int_{\ln 2}^{\ln 3} f(x) \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module