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B
13cos3x+c
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C
12sin2x+c
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D
13sin3x+c
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Solution
The correct option is D13sin3x+c ∫sin2xcosxdx
Let u=sinx dudx=cosx⇒du=cosxdx
By replacing sinx=u and cosx=du, we have ∫sin2cosxdx=∫u2du =u33+c =13sin3x+c