Mathematics

# Solve $\int\limits_0^{\dfrac{x}{2}} {\log \sin x\,dx}$

##### SOLUTION

We have,

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

We know that

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

Therefore,

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

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

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

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

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

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

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

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

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

Let $t=2x$

$dt=2dx$

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

Therefore,

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

We know that

$\int_{0}^{a}{f\left( x \right)}dx=2\int_{0}^{\dfrac{a}{2}}{f\left( x \right)}dx$

Therefore,

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

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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