int frac{dtheta}{sin theta cos^3 theta} = | vedclass

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Question

(C) Let $I = \int \frac{d	heta}{\sin 	heta \cos^3 	heta}$.Divide the numerator and denominator by $\cos^4 	heta$:$I = \int \frac{\sec^4 	heta \,

Vedclass · Accepted Answer

$\log 	an 	heta + \frac{1}{2} 	an^2 	heta + c$

Vedclass · Answer

(C) Let $I = \int \frac{d	heta}{\sin 	heta \cos^3 	heta}$.Divide the numerator and denominator by $\cos^4 	heta$:$I = \int \frac{\sec^4 	heta \, d	heta}{	an 	heta} = \int \frac{\sec^2 	heta (1 + 	an^2 	heta) \, d	heta}{	an 	heta}$.Substitute $t = 	an 	heta$,so $dt = \sec^2 	heta \, d	heta$:$I = \int \frac{1 + t^2}{t} \, dt = \int \left( \frac{1}{t} + t ight) \, dt$.Integrating with respect to $t$:$I = \log |t| + \frac{t^2}{2} + c$.Substituting back $t = 	an 	heta$:$I = \log |	an 	heta| + \frac{	an^2 	heta}{2} + c$.

$\int \frac{d\theta}{\sin \theta \cos^3 \theta} = $

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