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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
$\int\text{e}^{-3\text{x}}\cos^3\text{x dx}$
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Answer

Let $\text{I}=\int\text{e}^{-3\text{x}}\cos^3\text{x dx}$
$=\int\text{e}^{-3\text{x}}\Big(\frac{\cos3\text{x}+3\cos\text{x}}{4}\Big)\text{dx}$
$=\frac{1}{4}\int\big(\text{e}^{-3\text{x}}\cos3\text{x}+\text{e}^{-3\text{x}}\cos\text{x}\big)\text{dx}$
$=\frac{1}{4}(\text{I}_1+\text{I}_2)$
$\text{I}_1=\int\text{e}^{-3\text{x}}\cos3\text{x dx}$
$=\text{e}^{-3\text{x}}\int\cos3\text{x dx}-\int\big((\text{e}^{-3\text{x}})\int\cos3\text{x dx}\big)\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}-\int-3\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}+\int\text{e}^{-3\text{x}}\sin3\text{x dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}+\text{e}^{-3\text{x}}\int\sin3\text{x dx}-\int\big((\text{e}^{-3\text{x}})\int\sin3\text{x dx})\text{dx}$
$=\text{e}^{-3\text{x}}\frac{\sin3\text{x}}{3}-\text{e}^{-3\text{x}}\frac{\cos3\text{x}}{3}-\int\text{e}^{-3\text{x}}\cos3\text{x dx}$
$=\text{e}^{-3\text{x}}\frac{3\sin}{3}-\text{e}^{-3\text{x}}\frac{\cos3\text{x}}{3}-\text{I}_1$
$\Rightarrow\ 2\text{I}_1=\frac{\text{e}^{-3\text{x}}}{3}(\sin3\text{x}-\cos3\text{x})$
$\Rightarrow\ \text{I}_1=\frac{\text{e}^{-3\text{x}}}{6}(\sin3\text{x}-\cos3\text{x})+\text{C}_1$
Similarly $\text{I}_2=\int\text{e}^{-3\text{x}}\cos\text{x dx}=\frac{\text{e}^{-3\text{x}}}{10}(\sin3\text{x}-3\cos3\text{x})+\text{C}_2$
$\Rightarrow\ \text{I}=\frac{1}{4}\bigg[\frac{\text{e}^{-3\text{x}}}{6}(\sin3\text{x}-\cos3\text{x})+\frac{\text{e}^{-3\text{x}}}{10}(\sin3\text{x}-3\cos3\text{x})\bigg]+\text{C}$

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