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Question

The value of ∫√tanxsinxcosxdx is equal to

A
2tanx+C
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B
2cotx+C
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C
tanx2+C
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D
tanx+C
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Solution

The correct option is A 2√tanx+CGiven ∫√tanxsinxcosxsimplifying the function=∫√tanxsinx.cosx.cosxcosx=∫√tanxsinx.cos2xcosx=∫√tanxcos2x.sinxcosx=∫√tanxcos2x.tanx=∫√tanx.(tanx)−1cos2x=∫(tanx)1/2−1cos2x=∫(tanx)−1/2cos2x=∫(tanx)−1/2.1cos2x=∫(tanx)−1/2.sec2xLet tanx=tSo, sec2x=dtdx⇒dx=dtsec2x∴∫(tanx)−1/2.sec2x.dx=∫(t)−1/2.sec2x.dtsec2x=∫(t)−1/2dt=t−1/2+1−1/2+1+C {as∫xndx=xn+1n+1+C}=t1/21/2+C=2t1/2+C=2√t+CSubstituting t=tanx=2√tanx+C

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