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,

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

 $$ 2I=I-\dfrac{\pi }{2}\log 2 $$

 $$ I=-\dfrac{\pi }{2}\log 2 $$

Hence, this is the answer.

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