int frac{(1-4 sin^2 x) cos x}{cos (3x+2)} dx | vedclass

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(D) We have $I = \int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx$. Using the identity $1 - 4 \sin^2 x = 1 - 4(1 - \cos^2 x) = 4 \cos^2 x - 3$. So, $

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$(\cos 2) x + \frac{1}{3}(\sin 2) \log |\sec (3x+2)| + c$

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(D) We have $I = \int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx$. Using the identity $1 - 4 \sin^2 x = 1 - 4(1 - \cos^2 x) = 4 \cos^2 x - 3$. So, $I = \int \frac{(4 \cos^2 x - 3) \cos x}{\cos (3x+2)} dx = \int \frac{4 \cos^3 x - 3 \cos x}{\cos (3x+2)} dx$. Using the triple angle formula $\cos 3x = 4 \cos^3 x - 3 \cos x$, we get $I = \int \frac{\cos 3x}{\cos (3x+2)} dx$. Let $t = 3x+2$, then $dt = 3 dx$, so $dx = \frac{dt}{3}$. Also, $3x = t-2$. $I = \int \frac{\cos(t-2)}{\cos t} \cdot \frac{dt}{3} = \frac{1}{3} \int \frac{\cos t \cos 2 + \sin t \sin 2}{\cos t} dt$. $I = \frac{1}{3} \int (\cos 2 + \sin 2 	an t) dt = \frac{1}{3} [t \cos 2 + \sin 2 \ln |\sec t|] + c$. Substituting $t = 3x+2$, we get $I = \frac{1}{3} [(3x+2) \cos 2 + \sin 2 \ln |\sec (3x+2)|] + c = x \cos 2 + \frac{1}{3} \sin 2 \ln |\sec (3x+2)| + c'$.

$\int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx =$

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