int frac{d x}{4+5 cos x} = | vedclass

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(B) Let $I = \int \frac{dx}{4+5 \cos x}$.Using the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ and $dx = \frac{2 du}{1+u^2}$ where $u

Vedclass · Accepted Answer

$\frac{1}{3} \log \left|\frac{3+	an \frac{x}{2}}{3-	an \frac{x}{2}}ight|+C$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{4+5 \cos x}$.Using the substitution $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ and $dx = \frac{2 du}{1+u^2}$ where $u = 	an(x/2)$:$I = \int \frac{\frac{2 du}{1+u^2}}{4+5\left(\frac{1-u^2}{1+u^2}ight)} = \int \frac{2 du}{4(1+u^2) + 5(1-u^2)} = \int \frac{2 du}{4+4u^2+5-5u^2} = \int \frac{2 du}{9-u^2}$.Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} ight| + C$:$I = 2 	imes \frac{1}{2(3)} \ln \left| \frac{3+u}{3-u} ight| + C = \frac{1}{3} \ln \left| \frac{3+	an(x/2)}{3-	an(x/2)} ight| + C$.

$\int \frac{d x}{4+5 \cos x} = $

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