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

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