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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{1}{\text{x}^3}\sin(\log\text{x})\text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{\text{x}^3}\sin(\log\text{x})\text{dx}$
Putting $ \log x = t$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{e}^{3\text{t}}}\sin\text{t e}^\text{t}\text{dt}$
$=\int\text{e}^{-2\text{t}}\sin\text{t dt}$
Considering \sin $t$ as first function and $e^{-2t} $ as second function
$\text{I}=\sin\text{t}\Big[\frac{\text{e}^{-2\text{t}}}{-2}\Big]-\int\cos\text{t}\frac{\text{e}^{-2\text{t}}}{-2}\text{dt}$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}+\frac{1}{2}\int\cos\text{t e}^{-2\text{t}}\text{dt}$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}+\frac{1}{2}\Big[\cos\text{t}\frac{\text{e}^{-2\text{t}}}{-2}-\int(-\sin\text{t})\frac{\text{e}^{-2\text{t}}}{-2}\text{dt}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}-\frac{1}{4}\cos\text{t e}^{-2\text{t}}-\int\frac{\text{e}^{-2\text{t}}\sin\text{t dt}}{4}$
$\Rightarrow\text{I}=\text{e}^{-2\text{t}}\Big[\frac{-2\sin\text{t}-\cos\text{t}}{4}\Big]-\frac{\text{I}}{4}$
$\Rightarrow\frac{5\text{I}}{4}=\text{e}^{-2\text{t}}\Big[\frac{-2\sin\text{t}-\cos\text{t}}{4}\Big]$
$\Rightarrow\text{I}=\frac{\text{e}^{-2\text{t}}}{5}[-2\sin\text{t}-\cos\text{t}]+\text{C}$
$\Rightarrow\text{I}=\frac{-\text{x}^{-2}}{5}[2\sin(\log\text{x})+\cos(\log\text{x})]+\text{C}$
$\Rightarrow\text{I}=\frac{-1}{5\text{x}^2}[\cos(\log\text{x})+2\sin(\log\text{x})]+\text{C}$

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