If ∫ cosxdx ^ 3 x(1 + ^ 6 x)(2)/(3) = f(x)(1 + ^ 6 x)(1)/(λ) + c ,...

Let, $\sin x = t$

$cosxdx = dt$
$I = ∫ \dfrac{dt}{{t}^{3}\left(1 + {t}^{6}\right)\dfrac{2}{3}} = ∫ \dfrac{dt}{{t}^{7}\left(1 + \dfrac{1}{{t}^{6}}\right)\dfrac{2}{3}}$

Put $1 + \dfrac{1}{{t}^{6}} = {r}^{3}$

$⇒ \dfrac{dt}{{t}^{7}} = \dfrac{- 1}{2}{r}^{2}dr$

$I = - \dfrac{1}{2} ∫ \dfrac{{r}^{2}dr}{{r}^{2}} = - \dfrac{1}{2}r + c$

$⇒ I = - \dfrac{1}{2}\left(\dfrac{\sin^{6}x + 1}{\sin^{6}x}\right)\dfrac{1}{3} + c$

$⇒ I = - \dfrac{1}{2\sin^{2}x}\left(1 + \sin^{6}x\right)\dfrac{1}{3} + c$

So, $f\left(x\right) = - \dfrac{1}{2}cose{c}^{2}x$ and $\lambda = 3$

$\lambdaf\left(\dfrac{\pi}{3}\right) = - 2$