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Question

dθ(sinθ2cosθ)(2sinθ+cosθ)=

A
15log|tanθ22tanθ+1|+c
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B
12log|tanθ22tanθ+1|+c
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C
15log|2tanθ+1tanθ2|+c
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D
12log|2tanθ1tanθ+2|+c
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Solution

The correct option is A 15log|tanθ22tanθ+1|+c
Given, dθ(sinθ2cosθ)(2sinθ+cosθ)
Dividing numerator and denominator by cos2θ we have;
sec2θdθ(tanθ2)(2tanθ+1)
Breaking it in partial fraction form we have:
[15(sec2θtanθ2)25(sec2θ2tanθ+1)]dθ
15log(tanθ2)25×2log(2tanθ+1)+c
15log(tanθ2)(2tanθ+1)+c
Hence, option 'A' is correct.

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