Question

log cos2x ka definite intigral​

Answer 1

Given Integrand,

$\displaystyle \sf \: l = \int _{0}^{\frac{\pi}{2}} log(cos2x)dx - - - - - (1)$

We know that,

$\displaystyle \sf\int _{0}^{a} f(x)dx = \displaystyle \sf\int _{0}^{a} f(a - x)dx$

Thus,

$\longrightarrow \displaystyle \sf \: l = \int _{0}^{\frac{\pi}{2}} log(cos \big( \dfrac{\pi}{2} - 2x \big))dx \\ \\ \longrightarrow \displaystyle \sf \: l = \int _{0}^{\frac{\pi}{2}} log(sin2x)dx - - - - - - (2)$

Adding equations (1) and (2),

$\displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} log(cos2x)dx + \int _{0}^{\frac{\pi}{2}} log(sin2x)dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} \bigg[log(sin2x)+ log(cos2x) \bigg] dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} log(sin2x.cos2x)dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} log( \dfrac{2sin2xcos2x}{2} )dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} log( \dfrac{sin4x}{2})dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \int _{0}^{\frac{\pi}{2}} log(sin4x)dx - \int _{0}^{\frac{\pi}{2}}log(2)dx$

Let t = 4x, differentiating w.r.t x :

$\sf \: \dfrac{dt}{4} = dx$

When x = 0, t = 0

When x = π/2, t = 2π

$\longrightarrow \: \displaystyle \sf \: 2l = \dfrac{1}{4} \int _{0}^{2\pi} log(sint)dt - log(2)\int _{0}^{ \frac{\pi}{2} }dx \\ \\ \longrightarrow \: \displaystyle \sf \: 2l = \dfrac{1}{4} \int _{0}^{2\pi} log(sint)dt - \dfrac{\pi log(2) }{2} \\ \\ \longrightarrow \sf 2l = l - \dfrac{\pi log(2)}{2} \\ \\ \longrightarrow \boxed{\boxed{\sf l = - \dfrac{\pi log(2)}{2}}}$