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 If cosxdxsin3x1+sin6x2/3=f(x)1+sin6x1/λ+c where c is a constant of integration,  then λfπ3 is equal to: 

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a
−98
b
-2
c
2
d
98

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detailed solution

Correct option is B

Let sinx=t⇒cosxdx=dtNow I=∫dtt31+t623=∫dtt71+1t623 Put 1+1t6=r3⇒dtt7=-12r2dr⇒-12∫r2drr2=-12r+c=-12sin6x+1sin6x13+c=-12sin2x1+sin6x13+c⇒f(x)=-12cosec2x and λ=3 ∴λfπ3=-2


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