If int frac{dx}{5 + 4cos x} = lambda tan^{-1} | vedclass

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(A) We use the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.$\int \frac{dx}{5 + 4\cos x} = \int \frac{dx}{5 + 4\left( \frac{1 - 	a

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$\lambda = 2/3$

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(A) We use the substitution $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$.$\int \frac{dx}{5 + 4\cos x} = \int \frac{dx}{5 + 4\left( \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)} ight)}$$= \int \frac{(1 + 	an^2(x/2)) dx}{5(1 + 	an^2(x/2)) + 4(1 - 	an^2(x/2))}$$= \int \frac{\sec^2(x/2) dx}{9 + 	an^2(x/2)}$Let $	an(x/2) = y$,then $\frac{1}{2} \sec^2(x/2) dx = dy$,so $\sec^2(x/2) dx = 2 dy$.$= \int \frac{2 dy}{9 + y^2} = 2 \int \frac{dy}{3^2 + y^2} = 2 \cdot \frac{1}{3} 	an^{-1} \left( \frac{y}{3} ight) + C$$= \frac{2}{3} 	an^{-1} \left( \frac{1}{3} 	an \frac{x}{2} ight) + C$.Comparing with $\lambda 	an^{-1} (m 	an(x/2))$,we get $\lambda = 2/3$ and $m = 1/3$.

If $\int \frac{dx}{5 + 4\cos x} = \lambda \tan^{-1} \left( m \tan \frac{x}{2} \right) + C$,then:

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