Grade 12th passIntegral Calculus

integral from 0 to pi/2 logcosx ?

nirmala

11 Years agoGrade 12th pass

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$I = \int_{0}^{\frac{\pi }{2}}log(cosx)dx$ …..........(1)
$I = \int_{0}^{\frac{\pi }{2}}log(cos(\frac{\pi }{2}-x))dx$
$I = \int_{0}^{\frac{\pi }{2}}log(sinx)dx$ …...........(2)
(1) + (2)
$2I = \int_{0}^{\frac{\pi }{2}}(log(cosx)+ log(sinx))dx$
$2I = \int_{0}^{\frac{\pi }{2}}(log(sinx.cosx))dx$
$2I = \int_{0}^{\frac{\pi }{2}}(log(\frac{2}{2}sinx.cosx))dx$
$2I = \int_{0}^{\frac{\pi }{2}}(log({2}sinx.cosx)-log(2))dx$
$2I = \int_{0}^{\frac{\pi }{2}}log(sin2x)-\int_{0}^{\frac{\pi }{2}}log(2)dx$ …...(1)
$I_{1} = \int_{0}^{\frac{\pi }{2}}log(sin2x)dx$
$2x = t$
$2dx = dt$

$x = 0\rightarrow t = 0$
$x = \frac{\pi }{2}\rightarrow t = \pi$
$I_{1} = \frac{1}{2}\int_{0}^{\pi}log(sint)dt$
$I_{1} = \frac{1}{2}.2\int_{0}^{\pi/2}log(sint)dt$
$I_{1} = \int_{0}^{\pi/2}log(sint)dt$
$I_{1} = I$
Put in (1), we have
$I = -\int_{0}^{\frac{\pi }{2}}log(2)dx$
$I = -log(2)[x]_{0}^{\frac{\pi }{2}}$
$I = -\frac{\pi }{2}log(2)$