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Question

# Write a value of $\int {e}^{ax}\mathrm{cos}bxdx$.

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Solution

## $\mathrm{Let}I=\int {e}^{ax}.\mathrm{cos}bxdx\phantom{\rule{0ex}{0ex}}=\mathrm{cos}bx\int {e}^{ax}dx-\int \left\{\frac{d}{dx}\left(\mathrm{cos}bx\right)\int {e}^{ax}dx\right\}dx\phantom{\rule{0ex}{0ex}}=\mathrm{cos}bx×\frac{{e}^{ax}}{a}-\int -\mathrm{sin}bx×b.\frac{{e}^{ax}}{a}\phantom{\rule{0ex}{0ex}}=\mathrm{cos}bx×\frac{{e}^{ax}}{a}+\frac{b}{a}\int {e}^{ax}.\mathrm{sin}bxdx\phantom{\rule{0ex}{0ex}}=\mathrm{cos}bx×\frac{{e}^{ax}}{a}+\frac{b}{a}{I}_{1}...\left(1\right)\phantom{\rule{0ex}{0ex}}\therefore {I}_{1}=\int {e}^{ax}×\mathrm{sin}bxdx\phantom{\rule{0ex}{0ex}}=\mathrm{sin}bx\int {e}^{ax}dx-\int \left\{\frac{d}{dx}\left(\mathrm{sin}bx\right)\int {e}^{ax}dx\right\}dx\phantom{\rule{0ex}{0ex}}=\mathrm{sin}bx×\frac{{e}^{ax}}{a}-\int b.\mathrm{cos}bx×\frac{{e}^{ax}}{a}dx\phantom{\rule{0ex}{0ex}}=\mathrm{sin}bx.\frac{{e}^{ax}}{a}-\frac{b}{a}I\mathit{}....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)&\left(2\right)\phantom{\rule{0ex}{0ex}}\therefore I=\mathrm{cos}bx.\frac{{e}^{ax}}{a}+\frac{b}{a}\left\{\mathrm{sin}bx.\frac{{e}^{ax}}{a}-\frac{b}{a}I\right\}\phantom{\rule{0ex}{0ex}}⇒I=\mathrm{cos}bx.\frac{{e}^{ax}}{a}+\frac{b}{{a}^{2}}\mathrm{sin}bx{e}^{ax}-\frac{{b}^{2}}{{a}^{2}}I\phantom{\rule{0ex}{0ex}}⇒I+\frac{{b}^{2}}{{a}^{2}}I=\mathrm{cos}bx.\frac{{e}^{ax}}{a}+\frac{b\mathrm{sin}bx{e}^{ax}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}⇒\left({a}^{2}+{b}^{2}\right)I=\left(a\mathrm{cos}bx+b\mathrm{sin}bx\right){e}^{ax}\phantom{\rule{0ex}{0ex}}⇒I=\frac{\left(a\mathrm{cos}bx+b\mathrm{sin}bx\right){e}^{ax}}{{a}^{2}+{b}^{2}}+C$

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