Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int\cos(\log\text{x})\text{dx}$
✓
Answer
Let $\text{I}=\int\cos(\log\text{x})\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{(t)dt}$
Considering cos (t) as first function and $e^t$ as second function
$\text{I}=\cos\text{t}\text{e}^\text{t}-\int(-\sin\text{t})\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{t}\text{e}^\text{t}+\int\sin\text{t}\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{te}^\text{t}+\text{I}_1\ \dots(1)$
where $\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
$\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
Considering sin t as first function and $e^t$ as second function
$\text{I}_1=\sin\text{te}^\text{t}-\int\cos\text{te}^\text{t}\text{dt}$
$\Rightarrow\text{I}_1=\sin\text{te}^\text{t}-\text{I}\ \dots(2)$
From (1) & (2)
$\text{I}=\cos\text{te}^\text{t}+\sin\text{te}^\text{t}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}(\sin\text{t}+\cos\text{t})}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{e}^{\log\text{x}}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{x}}{2}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]+\text{C}$
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