# Prove that $$\int_{0}^{\pi/2}$$ log(sinx)dx = $$\int_{0}^{\pi/2}$$ log(cosx)dx = -$$\pi\over 2$$log 2.

## Solution :

Let I = $$\int_{0}^{\pi/2}$$ log(sinx)dx    …….(i)

then I = $$\int_{0}^{\pi/2}$$ $$log(sin({\pi\over 2}-x))$$dx = $$\int_{0}^{\pi/2}$$ log(cosx)dx     …….(ii)

Adding (i) and (ii), we get

2I = $$\int_{0}^{\pi/2}$$ log(sinx)dx + $$\int_{0}^{\pi/2}$$ log(cosx)dx = $$\int_{0}^{\pi/2}$$ (log(sinx)dx + log(cosx))

$$\implies$$ $$\int_{0}^{\pi/2}$$ log(sinxcosx)dx = $$\int_{0}^{\pi/2}$$ $$log({2sinxcosx\over 2})$$dx

= $$\int_{0}^{\pi/2}$$ $$log({sin2x\over 2})$$dx = $$\int_{0}^{\pi/2}$$ log(sin2x)dx – $$\int_{0}^{\pi/2}$$ log(2)dx

= $$\int_{0}^{\pi/2}$$ log(sin2x)dx – (log 2)$${(x)}^{\pi/2}_{0}$$

$$\implies$$  2I = $$\int_{0}^{\pi/2}$$ log(sin2x)dx – $$\pi\over 2$$log 2   …(iii)

Let $$I_1$$ = $$\int_{0}^{\pi/2}$$ log(sin2x)dx,  putting 2x = t, we get

$$I_1$$ = $$\int_{0}^{\pi}$$ log(sint)$$dt\over 2$$ = $$1\over 2$$ $$\int_{0}^{\pi}$$ log(sint)dt = $$1\over 2$$ 2$$\int_{0}^{\pi/2}$$ log(sint)dt

$$I_1$$ = $$\int_{0}^{\pi/2}$$ log(sinx)dx

$$\therefore$$  (iii) becomes; 2I = I – $$\pi\over 2$$log 2

Hence  $$\int_{0}^{\pi/2}$$ log(sinx)dx = – $$\pi\over 2$$log 2

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