∫sec⁡(x−5)sin⁡(x−3)dx=1cos⁡2log⁡(f(x))+C then f(x)=

I=∫sec⁡(x−5)sin⁡(x−3)dx=1cos⁡(2)∫cos⁡((x−5)−(x−3))sin⁡(x−3)cos⁡(x−5)dx=1cos⁡2∫cos⁡(x−5)cos⁡(x−3)+sin⁡(x−5)sin⁡(x−3)sin⁡(x−3)cos⁡(x−5)dx=1cos⁡2∫cot⁡(x−3)+tan⁡(x−5)dx=1cos⁡2[log⁡(sin⁡(x−3))+log⁡(sec⁡(x−5))]+C=1cos⁡2log⁡sin⁡(x−3)cos⁡(x−5)+C

∫sec⁡(x−5)sin⁡(x−3)dx=1cos⁡2log⁡(f(x))+C then f(x)=

I=∫sec⁡(x−5)sin⁡(x−3)dx=1cos⁡(2)∫cos⁡((x−5)−(x−3))sin⁡(x−3)cos⁡(x−5)dx=1cos⁡2∫cos⁡(x−5)cos⁡(x−3)+sin⁡(x−5)sin⁡(x−3)sin⁡(x−3)cos⁡(x−5)dx=1cos⁡2∫cot⁡(x−3)+tan⁡(x−5)dx=1cos⁡2[log⁡(sin⁡(x−3))+log⁡(sec⁡(x−5))]+C=1cos⁡2log⁡sin⁡(x−3)cos⁡(x−5)+C

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$\int \frac{\sec (x - 5)}{\sin (x - 3)} d x = \frac{1}{\cos 2} \log (f (x)) + C then f (x) =$

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$\frac{\sin (x - 3)}{\cos (x - 5)}$

$\frac{\cos (x - 5)}{\cos (x - 3)}$

$\frac{\cos (x - 3)}{\cos (x - 5)}$

$\frac{\sin (x - 5)}{\cos (x - 3)}$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{\sec (x - 5)}{\sin (x - 3)} d x = \frac{1}{\cos (2)} \int \frac{\cos ((x - 5) - (x - 3))}{\sin (x - 3) \cos (x - 5)} d x \\ = \frac{1}{\cos 2} \int \frac{\cos (x - 5) \cos (x - 3) + \sin (x - 5) \sin (x - 3)}{\sin (x - 3) \cos (x - 5)} d x \end{array}$

$\begin{array}{l} = \frac{1}{\cos 2} \int \cot (x - 3) + \tan (x - 5) d x \\ = \frac{1}{\cos 2} [\log (\sin (x - 3)) + \log (\sec (x - 5))] + C \\ = \frac{1}{\cos 2} [\log (\frac{\sin (x - 3)}{\cos (x - 5)})] + C \end{array}$