Mathematics

$$\displaystyle \int^{\pi/2}_{0}In (\sin x)dx$$ equals


ANSWER

$$ -\dfrac{\pi }{2}\ln 2 $$


SOLUTION

Consider the given integral.

$$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)dx}$$                   …….. (1)

 

We know that

$$\int_{a}^{b}{f\left( x \right)dx=}\int_{a}^{b}{f\left( a+b-x \right)dx}$$

 

Therefore,

$$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin \left( \dfrac{\pi }{2}-x \right) \right)dx}$$

$$I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \cos x \right)dx}$$                            ……. (2)

 

On adding equation (1) and (2), we get

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x \right)+\ln \left( \cos x \right)dx} $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin x\cos x \right)dx} $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \dfrac{2\sin x\cos x}{2} \right)dx} $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\int_{0}^{\dfrac{\pi }{2}}{\ln \left( 2 \right)dx} $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\ln 2\left( x \right)_{0}^{\dfrac{\pi }{2}} $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin 2x \right)dx}-\dfrac{\pi }{2}\ln 2 $$

 

Let $$t=2x$$

$$\dfrac{dt}{2}=dx$$

 

Therefore,

$$ 2I=\dfrac{1}{2}\int_{0}^{\pi }{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2 $$

$$ 2I=\dfrac{2}{2}\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \int_{0}^{a}{f\left( x \right)dx=2\int_{0}^{a/2}{f\left( x \right)dx}} \right] $$

$$ 2I=\int_{0}^{\dfrac{\pi }{2}}{\ln \left( \sin t \right)dt}-\dfrac{\pi }{2}\ln 2 $$

$$ 2I=I-\dfrac{\pi }{2}\ln 2 $$

$$ I=-\dfrac{\pi }{2}\ln 2 $$

 

Hence, this is the answer.

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