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Question

Let α∈(0,π2) be fixed. If the integral ∫tanx+tanαtanx−tanαdx=A(x)cos2α+B(x)sin2α+C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

A
xα and loge|sin(xα)|
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B
x+α and loge|sin(xα)|
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C
xα and loge|cos(xα)|
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D
x+α and loge|sin(x+α)|
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Solution

The correct option is A x−α and loge|sin(x−α)|∫tanx+tanαtanx−tanαdx ⇒∫sinxcosα+cosxsinαsinxcosα−cosxsinαdx=∫sin(x+α)sin(x−α)dx=∫sin(x−α+2α)sin(x−α)dx=∫sin(x−α)cos2αsin(x−α)dx+∫cos(x−α)sin2αsin(x−α)dx=xcos2α+sin2αlog|sin(x−α)|+c Here, option (a) and (b) both are correct so it is bonus.

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