If frac{3 pi}{4}

Question

(C) Given the integral $I = \int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx$. Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2

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$\frac{2^x}{\log 2} + \sin x - \cos x - \frac{1}{x} - \log |x| + c$

Vedclass · Answer

(C) Given the integral $I = \int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx$. Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$, we have $\sqrt{1 + \sin 2x} = |\sin x + \cos x|$. Given $\frac{3 \pi}{4} Thus, $|\sin x + \cos x| = -(\sin x + \cos x)$. Substituting this into the integral: $I = \int \left(2^x - (-(\sin x + \cos x)) + \frac{1}{x^2} - \frac{1}{x}\right) dx$ $I = \int \left(2^x + \sin x + \cos x + \frac{1}{x^2} - \frac{1}{x}\right) dx$ Integrating term by term: $I = \frac{2^x}{\log 2} - \cos x + \sin x - \frac{1}{x} - \log |x| + c$ Rearranging the terms, we get: $I = \frac{2^x}{\log 2} + \sin x - \cos x - \frac{1}{x} - \log |x| + c$. Therefore, option $C$ is correct.

If $\frac{3 \pi}{4} < x < \frac{7 \pi}{4}$, then $\int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx = $

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