int e^{tan theta} (sec theta - sin theta) dth | vedclass

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Question

(D) We have $I = \int e^{	an 	heta} (\sec 	heta - \sin 	heta) d	heta$. Rewrite the integrand: $I = \int e^{	an 	heta} \sec 	heta (1 - \sin 	h

Vedclass · Accepted Answer

$e^{	an 	heta} \cos 	heta + c$

Vedclass · Answer

(D) We have $I = \int e^{	an 	heta} (\sec 	heta - \sin 	heta) d	heta$. Rewrite the integrand: $I = \int e^{	an 	heta} \sec 	heta (1 - \sin 	heta \cos 	heta) d	heta$. Actually,let us simplify the expression: $I = \int e^{	an 	heta} \sec 	heta d	heta - \int e^{	an 	heta} \sin 	heta d	heta$. Using the identity $\int e^{f(x)} (f'(x) g(x) + g'(x)) dx = e^{f(x)} g(x) + c$. Let $f(	heta) = 	an 	heta$,then $f'(	heta) = \sec^2 	heta$. $I = \int e^{	an 	heta} \sec 	heta d	heta - \int e^{	an 	heta} \sin 	heta d	heta$. Note that $\int e^{	an 	heta} \sec 	heta d	heta$ is not standard,but we can rewrite the original integral as: $I = \int e^{	an 	heta} \sec 	heta (1 - \sin 	heta \cos 	heta) d	heta = \int e^{	an 	heta} (\sec 	heta - 	an 	heta \sin 	heta) d	heta$. This does not simplify easily. Let us re-evaluate: $I = \int e^{	an 	heta} \sec 	heta d	heta - \int e^{	an 	heta} \sin 	heta d	heta$. Actually,the standard form is $\int e^{	an 	heta} (\sec 	heta + \sec 	heta 	an 	heta) d	heta = e^{	an 	heta} \sec 	heta + c$. Given the options,let us check the derivative of $e^{	an 	heta} \cos 	heta$: $\frac{d}{d	heta} (e^{	an 	heta} \cos 	heta) = e^{	an 	heta} \sec^2 	heta \cos 	heta + e^{	an 	heta} (-\sin 	heta) = e^{	an 	heta} (\sec 	heta - \sin 	heta)$. Thus,$\int e^{	an 	heta} (\sec 	heta - \sin 	heta) d	heta = e^{	an 	heta} \cos 	heta + c$.

$\int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta$ equals:

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