Mathematics

$$\int _{ 0 }^{ \pi /2 }{ x\sin { x } \cos { x } dx } $$


SOLUTION

Consider the following question.

$$ I=\int_{0}^{\dfrac{\pi }{2}}{x\sin x\cos xdx} $$

Using a Ilate rule, we get.

Therefore,

$$ =\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{x\sin 2xdx} $$

$$ =\dfrac{1}{2}\int_{0}^{\dfrac{\pi }{2}}{x\sin 2xdx} $$

$$ =\dfrac{1}{2}\left( \dfrac{1}{2}x\left( -\cos 2x \right)+\dfrac{1}{4}\sin 2x \right)_{0}^{\dfrac{\pi }{2}} $$

$$ =\dfrac{1}{2}\left( x\left( -\cos 2x \right)+\dfrac{1}{2}\sin 2x \right)_{0}^{\dfrac{\pi }{2}} $$

$$ =\dfrac{1}{2}\left( \dfrac{\pi }{2}\left( -\cos 2\dfrac{\pi }{2} \right)+\dfrac{1}{2}\sin 2\dfrac{\pi }{2} \right)-\left( 0\left( -\cos 2\times 0 \right)+\dfrac{1}{2}\sin 2\times 0 \right) $$

$$ =\dfrac{\pi }{4} $$

Hence, this is the required answer.

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