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integral from 0 to pi/2 logcosx ?

integral from 0 to pi/2 logcosx ?
 

Grade:12th pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 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)

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