int frac{1}{1+3 sin ^2 x+8 cos ^2 x} d x is e | vedclass

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(B) Let $I = \int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$. Dividing the numerator and denominator by $\cos ^2 x$,we get: $I = \int \frac{\sec ^2 x}{\

Vedclass · Accepted Answer

$\frac{1}{6} 	an ^{-1}\left(\frac{2 	an x}{3}ight)+C$

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(B) Let $I = \int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$. Dividing the numerator and denominator by $\cos ^2 x$,we get: $I = \int \frac{\sec ^2 x}{\sec ^2 x+3 	an ^2 x+8} d x$. Since $\sec ^2 x = 1 + 	an ^2 x$,we have: $I = \int \frac{\sec ^2 x}{1 + 	an ^2 x + 3 	an ^2 x + 8} d x = \int \frac{\sec ^2 x}{4 	an ^2 x + 9} d x$. Let $	an x = t$,then $\sec ^2 x d x = d t$. Substituting these into the integral: $I = \int \frac{d t}{4 t^2 + 9} = \frac{1}{4} \int \frac{d t}{t^2 + (3/2)^2}$. Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = \frac{1}{4} 	imes \frac{1}{3/2} 	an^{-1}(\frac{t}{3/2}) + C = \frac{1}{4} 	imes \frac{2}{3} 	an^{-1}(\frac{2t}{3}) + C$. $I = \frac{1}{6} 	an^{-1}(\frac{2 	an x}{3}) + C$.

$\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$ is equal to

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