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

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(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: $\frac{\pi}{4} - \ln \sqrt{2}$.

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

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