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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Write a value of $\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
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Answer

Let $\text{I}=\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
$=\cos\text{bx}\int\text{e}^{\text{ax}}\text{ dx}-\Big\{\frac{\text{d}}{\text{dx}}(\cos\text{bx})\int\text{e}^{\text{ax}}\text{ dx}\Big\}\text{dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int-\sin\text{bx}\cdot\text{b}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\text{I}_1\ ....(\text{i})$
$\therefore\ \text{I}_1=\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\sin\text{bx}\int\text{e}^{\text{ax}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\sin\text{bx})\int\text{e}^{\text{ax}}\text{dx}\Big\}\text{dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int\text{b}\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}\text{ dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\ ....(\text{ii})$
From (i) & (ii)
$\therefore\ \text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\Big\{\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\Big\}$
$\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}^2}\sin\text{bx}\text{e}^{\text{ax}}-\frac{\text{b}^2}{\text{a}^2}\text{I}$
$\text{I}+\frac{\text{b}^2}{\text{a}^2}\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}\sin\text{bx}\text{e}^{\text{ax}}}{\text{a}^2}$
$\big(\text{a}^2+\text{b}^2\big)\text{I}=(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}$
$\text{I}=\frac{(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}}{\text{a}^2+\text{b}^2}+\text{C}$

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