int frac{1}{16-7 sin ^2 x} d x= | vedclass

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(A) To evaluate the integral $I = \int \frac{1}{16-7 \sin^2 x} dx$, divide the numerator and denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x}{16 \s

Vedclass · Accepted Answer

$\frac{1}{12} 	an^{-1}\left(\frac{3 	an x}{4}ight)+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{1}{16-7 \sin^2 x} dx$, divide the numerator and denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x}{16 \sec^2 x - 7 	an^2 x} dx$Using $\sec^2 x = 1 + 	an^2 x$, we get:$I = \int \frac{\sec^2 x}{16(1 + 	an^2 x) - 7 	an^2 x} dx = \int \frac{\sec^2 x}{16 + 9 	an^2 x} dx$Let $	an x = t$, then $\sec^2 x dx = dt$:$I = \int \frac{dt}{16 + 9t^2} = \frac{1}{9} \int \frac{dt}{\frac{16}{9} + t^2} = \frac{1}{9} \int \frac{dt}{(\frac{4}{3})^2 + t^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = \frac{1}{9} 	imes \frac{1}{4/3} 	an^{-1}\left(\frac{t}{4/3}ight) + C = \frac{1}{9} 	imes \frac{3}{4} 	an^{-1}\left(\frac{3t}{4}ight) + C$$I = \frac{1}{12} 	an^{-1}\left(\frac{3 	an x}{4}ight) + C$

$\int \frac{1}{16-7 \sin ^2 x} d x=$

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