Mathematics

Evaluate the following integrals:
$$\displaystyle \int { \cfrac { 1 }{ { x }^{ 3 } }  } \sin { \left( \log { x }  \right)  } dx\quad $$


SOLUTION
Let $$I=\displaystyle\int \dfrac{1}{x^3}\sin (\log x)d x$$

Put $$x=e^{t}\implies d x=e^t d t$$

$$\implies I=\displaystyle\int \dfrac{1}{(e^t)^3}\sin (\log e^t)\times e^t\ d t$$

$$\implies I=\displaystyle\int e^{-2 t}\sin t\  d t$$

As we know that

$$\displaystyle\int f(x) g (x) d x=f(x)\int g(x) d x-\int f'(x)\left(\int g(x) d x\right) d x$$

Here $$f(t)=\sin t,g(t)=e^{-2 t}$$

$$\implies f'(t)=\cos t$$    

$$\implies \displaystyle\int g(t) d t=\int e^{-2 t} d t=\dfrac{e^{-2 t}}{-2}=-\dfrac{1}{2}e^{-2 t}$$

$$\implies I=\sin t\times \bigg(\dfrac{-1}{2}e^{-2 t}\bigg)-\displaystyle\int \cos t\times \bigg(\dfrac{-1}{2}e^{-2 t}\bigg)d t$$

$$\implies I=-\dfrac{1}{2}e^{-2 t}\sin t+\dfrac{1}{2}\displaystyle\int e^{-2 t}\cos t\ d t$$

As we know that

$$\displaystyle\int f(x) g (x) d x=f(x)\int g(x) d x-\int f'(x)\left(\int g(x) d x\right) d x$$

Here $$f(t)=\cos t,g(t)=e^{-2 t}$$

$$\implies f'(t)=-\sin t$$    

$$\implies \displaystyle\int g(t) d t=\int e^{-2 t} d t=\dfrac{e^{-2 t}}{-2}=-\dfrac{1}{2}e^{-2 t}$$

$$\implies I=-\dfrac{1}{2}e^{-2 t}\sin t+\dfrac{1}{2}\left(\cos t\times \bigg(-\dfrac{1}{2}e^{-2 t}\bigg)-\displaystyle\int (-\sin t)\times \bigg(-\dfrac{1}{2}e^{-2 t}\bigg) d t\right)$$

$$\implies I=-\dfrac{1}{2}e^{-2 t}\sin t-\dfrac{1}{4}e^{-2 t}\cos t-\dfrac{1}{4}\displaystyle\int e^{-2 t}\sin t\  dt$$

$$\implies I=-\dfrac{1}{2}e^{-2 t}\sin t-\dfrac{1}{4}e^{-2 t}\cos t-\dfrac{1}{4}I$$

$$\implies I\bigg(1+\dfrac{1}{4}\bigg)=-\dfrac{1}{4}e^{-2 t}(2\sin t+\cos t)$$

$$\implies \dfrac{5}{4} I=-\dfrac{1}{4}e^{-2 t}(2\sin t+\cos t)$$

$$\implies  I=-\dfrac{1}{5 (e^{t})^2}(2\sin t+\cos t)$$

$$\implies I=-\dfrac{1}{5 x^2}(2\sin (\log x)+\cos (\log x))$$
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Subjective Medium Published on 17th 09, 2020
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