यदि int frac{x}{x tan x+1} , dx = log f(x) + | vedclass

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(C) माना $I = \int \frac{x}{x 	an x + 1} \, dx = \int \frac{x \cos x}{x \sin x + \cos x} \, dx$. माना $u = x \sin x + \cos x$. तब $du = (\sin x + x \

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$\frac{\pi + 4}{4 \sqrt{2}}$

Vedclass · Answer

(C) माना $I = \int \frac{x}{x 	an x + 1} \, dx = \int \frac{x \cos x}{x \sin x + \cos x} \, dx$. माना $u = x \sin x + \cos x$. तब $du = (\sin x + x \cos x - \sin x) \, dx = x \cos x \, dx$. अतः,$I = \int \frac{1}{u} \, du = \log |u| + k = \log |x \sin x + \cos x| + k$. इसकी तुलना $\log f(x) + k$ से करने पर,हमें $f(x) = |x \sin x + \cos x|$ प्राप्त होता है। अब,$f\left(\frac{\pi}{4}ight) = |\frac{\pi}{4} \sin\left(\frac{\pi}{4}ight) + \cos\left(\frac{\pi}{4}ight)|$ का मान ज्ञात करें। $f\left(\frac{\pi}{4}ight) = |\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}| = |\frac{\pi + 4}{4 \sqrt{2}}| = \frac{\pi + 4}{4 \sqrt{2}}$.

यदि $\int \frac{x}{x \tan x+1} \, dx = \log f(x) + k$ है,तो $f\left(\frac{\pi}{4}\right) =$

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