int frac{cos 2x}{cos x} dx = | vedclass

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Question

(C) We know that $\cos 2x = 2 \cos^2 x - 1$. Substituting this into the integral, we get: $\int \frac{2 \cos^2 x - 1}{\cos x} dx = \int (2 \cos x - \f

Vedclass · Accepted Answer

$2 \sin x - \log |\sec x + 	an x| + C$

Vedclass · Answer

(C) We know that $\cos 2x = 2 \cos^2 x - 1$. Substituting this into the integral, we get: $\int \frac{2 \cos^2 x - 1}{\cos x} dx = \int (2 \cos x - \frac{1}{\cos x}) dx$. This simplifies to $\int (2 \cos x - \sec x) dx$. Integrating term by term, we get: $2 \int \cos x dx - \int \sec x dx$. The integral of $\cos x$ is $\sin x$ and the integral of $\sec x$ is $\log |\sec x + 	an x|$. Thus, the final result is $2 \sin x - \log |\sec x + 	an x| + C$.

$\int \frac{\cos 2x}{\cos x} dx = $

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