int frac{1}{9 cos ^2 x-24 sin x cos x+16 sin | vedclass

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Question

(A) The given integral is $I = \int \frac{1}{9 \cos^2 x - 24 \sin x \cos x + 16 \sin^2 x} dx$. Divide the numerator and denominator by $\cos^2 x$: $I

Vedclass · Accepted Answer

$\frac{\cos x}{4(3 \cos x-4 \sin x)}+c$

Vedclass · Answer

(A) The given integral is $I = \int \frac{1}{9 \cos^2 x - 24 \sin x \cos x + 16 \sin^2 x} dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{9 - 24 	an x + 16 	an^2 x} dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. The integral becomes $I = \int \frac{du}{(3 - 4u)^2} = \int (3 - 4u)^{-2} du$. Using the power rule for integration: $I = \frac{(3 - 4u)^{-1}}{(-1) 	imes (-4)} + c = \frac{1}{4(3 - 4u)} + c$. Substitute $u = 	an x$ back: $I = \frac{1}{4(3 - 4 	an x)} + c = \frac{1}{4(3 - 4 \frac{\sin x}{\cos x})} + c = \frac{\cos x}{4(3 \cos x - 4 \sin x)} + c$. Thus,the correct option is $A$.

$\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x=$

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