int frac{dx}{4sin^2 x + 5cos^2 x} = | vedclass

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(C) To evaluate the integral $I = \int \frac{dx}{4\sin^2 x + 5\cos^2 x}$,divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{5}} 	an^{-1} \left( \frac{2	an x}{\sqrt{5}} ight) + c$

Vedclass · Answer

(C) To evaluate the integral $I = \int \frac{dx}{4\sin^2 x + 5\cos^2 x}$,divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x dx}{4	an^2 x + 5}$. Let $	an x = t$,then $\sec^2 x dx = dt$. The integral becomes $I = \int \frac{dt}{4t^2 + 5} = \frac{1}{4} \int \frac{dt}{t^2 + (\frac{\sqrt{5}}{2})^2}$. Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$,we get: $I = \frac{1}{4} \cdot \frac{1}{\frac{\sqrt{5}}{2}} 	an^{-1} \left( \frac{t}{\frac{\sqrt{5}}{2}} ight) + c = \frac{1}{2\sqrt{5}} 	an^{-1} \left( \frac{2t}{\sqrt{5}} ight) + c$. Substituting $t = 	an x$,we get $I = \frac{1}{2\sqrt{5}} 	an^{-1} \left( \frac{2	an x}{\sqrt{5}} ight) + c$.

$\int \frac{dx}{4\sin^2 x + 5\cos^2 x} = $

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