Mathematics

Integrate:
$$\displaystyle \int_{0}^{\pi }{\left( \sin mx\cdot \cos nx \right)}dx$$, where $$(m \ne n, \in to\,\, integer)$$


ANSWER

$$0$$


SOLUTION

Consider the given integral.

$$I=\int_{0}^{\pi }{\left( \sin mx\cdot \cos nx \right)}dx$$

 We know that

$$\sin A\cdot \sin B=\dfrac{1}{2}\left[ \cos \left( A-B \right)-\cos \left( A+B \right) \right]$$

 Therefore,

$$ I=\int_{0}^{\pi }{\left( \dfrac{1}{2}\left( \cos \left( mx-nx \right)-\cos \left( mx+nx \right) \right) \right)}dx $$

$$ I=\dfrac{1}{2}\int_{0}^{\pi }{\left( \cos \left( m-n \right)x-\cos \left( m+n \right)x \right)}dx $$

$$ I=\dfrac{1}{2}\int_{0}^{\pi }{\left( \cos \left( m-n \right)x \right)}dx-\dfrac{1}{2}\int_{0}^{\pi }{\left( \cos \left( m+n \right)x \right)}dx $$

$$ I=\dfrac{1}{2}\left[ \dfrac{\sin \left( m-n \right)x}{m-n} \right]_{0}^{\pi }-\dfrac{1}{2}\left[ \dfrac{\sin \left( m+n \right)x}{m+n} \right]_{0}^{\pi } $$

$$ I=\dfrac{1}{2}\left[ \dfrac{\sin \left( m-n \right)\pi }{m-n}-\dfrac{\sin \left( m-n \right)0}{m-n} \right]-\dfrac{1}{2}\left[ \dfrac{\sin \left( m+n \right)\pi }{m+n}-\dfrac{\sin \left( m+n \right)0}{m+n} \right] $$

$$ I=0 $$

 Hence, this is the answer.

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Single Correct Medium Published on 17th 09, 2020
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