Mathematics

If $$\int _{ 0 }^{ \pi  }{ \frac { \sin { x }  }{ { x }^{ 2 } }  } dx,\quad then\quad \int _{ 0 }^{ { \pi  }/{ 2 } }{ \frac { \cos { 2x }  }{ x }  } dx$$ is equal to :


ANSWER

$$1+A$$


SOLUTION
Given,

$$A=\int _0^{\pi }\:\:\dfrac{sinx}{x^2} dx$$

integration by parts $$u=\sin \left(x\right),\:v'=\dfrac{1}{x^2}$$

$$=\left[\dfrac{-\sin \left(x\right)-x\cdot \int \:-\frac{\cos \left(x\right)}{x}dx}{x}\right]^{\pi }_0$$

$$=\left[\dfrac{-\sin \left(x\right)-x\left(-\text{Ci}\left(x\right)\right)}{x}\right]^{\pi }_0$$

$$=\left[\dfrac{-\sin \left(x\right)+x\text{Ci}\left(x\right)}{x}\right]^{\pi }_0$$

$$=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$$

$$A=\text{Ci}\left(\pi \right)+1-\text{Ci}\left(0\right)$$

Now,

$$\int _0^{\frac{\pi }{2}}\dfrac{\cos \left(2x\right)}{x}dx$$

substitute $$u=2x$$

$$=\int _0^{\pi }\dfrac{\cos \left(u\right)}{u}du$$

$$=\left[\text{Ci}\left(u\right)\right]^{\pi }_0$$

$$=\text{Ci}\left(\pi \right)-\text{Ci}\left(0\right)$$

$$\therefore \int _0^{\frac{\pi }{2}}\dfrac{\cos \left(2x\right)}{x}dx=A+1$$

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Single Correct Medium Published on 17th 09, 2020
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