int frac{dx}{1 + 3sin^2 x} = | vedclass

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Question

(B) To evaluate the integral $I = \int \frac{dx}{1 + 3\sin^2 x}$,we divide the numerator and denominator by $\cos^2 x$ or rewrite the denominator usin

Vedclass · Accepted Answer

$\frac{1}{2}	an^{-1}(2	an x) + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{1 + 3\sin^2 x}$,we divide the numerator and denominator by $\cos^2 x$ or rewrite the denominator using the identity $1 = \sin^2 x + \cos^2 x$. $I = \int \frac{dx}{\sin^2 x + \cos^2 x + 3\sin^2 x} = \int \frac{dx}{4\sin^2 x + \cos^2 x}$. Divide numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x dx}{4	an^2 x + 1} = \frac{1}{4} \int \frac{\sec^2 x dx}{	an^2 x + \frac{1}{4}}$. Let $t = 	an x$,then $dt = \sec^2 x dx$. $I = \frac{1}{4} \int \frac{dt}{t^2 + (\frac{1}{2})^2}$. Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = \frac{1}{4} \cdot \frac{1}{1/2} 	an^{-1}(\frac{t}{1/2}) + c = \frac{1}{4} \cdot 2 	an^{-1}(2t) + c = \frac{1}{2} 	an^{-1}(2	an x) + c$.

$\int \frac{dx}{1 + 3\sin^2 x} = $

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