Evaluate int_{sin theta}^{cos theta} f(x tan | vedclass

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Question

(D) Let $u = x 	an 	heta$. Then $du = 	an 	heta \, dx$,which implies $dx = \frac{du}{	an 	heta} = \cot 	heta \, du$.When $x = \sin 	heta$,$u =

Vedclass · Accepted Answer

$\cot 	heta \int_{\sin 	heta}^{\sin 	heta 	an 	heta} f(x) \, dx$

Vedclass · Answer

(D) Let $u = x 	an 	heta$. Then $du = 	an 	heta \, dx$,which implies $dx = \frac{du}{	an 	heta} = \cot 	heta \, du$.When $x = \sin 	heta$,$u = \sin 	heta 	an 	heta$.When $x = \cos 	heta$,$u = \cos 	heta 	an 	heta = \sin 	heta$.Substituting these into the integral:$\int_{\sin 	heta}^{\cos 	heta} f(x 	an 	heta) \, dx = \int_{\sin 	heta 	an 	heta}^{\sin 	heta} f(u) \cot 	heta \, du$$= \cot 	heta \int_{\sin 	heta 	an 	heta}^{\sin 	heta} f(u) \, du$$= -\cot 	heta \int_{\sin 	heta}^{\sin 	heta 	an 	heta} f(u) \, du$.Comparing this with the given options,the correct form is represented by option $D$.

Evaluate $\int_{\sin \theta}^{\cos \theta} f(x \tan \theta) \, dx$,where $\theta \neq \frac{n \pi}{2}, n \in I$.

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