What is the integral of $$\int e^{ax}.sin(bx)dx$$ I used LIATE rule but even after that I am going round and round but not getting the answer.
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2Hint: this is $$\Im\left(\int e^{(a+ib)x}dx\right).$$ – Did Mar 09 '16 at 15:13
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1Please state LIATE rule in the question. – coffeemath Mar 09 '16 at 15:14
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Can you give one more hint – Archis Welankar Mar 09 '16 at 15:16
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Logarithmic ,inverse,algebraic,trigo,exponential i thought its well known. – Archis Welankar Mar 09 '16 at 15:17
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4Duplicate: $\displaystyle\int e^{ax}\sin(bx),\mathrm{d}x$. See also this thread in the Integrals and Series forum detailing three different ways to evaluate that integral. – Workaholic Mar 09 '16 at 15:33
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$$ du = \mathrm{e}^{ax}, u = \frac{1}{a}\mathrm{e}^{ax}\\ v = \sin (bx), dv = b\cos (bx) $$ then we get $$ I = \int \mathrm{e}^{ax}\sin (bx) dx = \frac{1}{a}\mathrm{e}^{ax}\sin (bx) - \frac{b}{a}\int \mathrm{e}^{ax}\cos (bx)dx $$ then we have $$ \int \mathrm{e}^{ax}\cos (bx)dx $$ using by parts again $$ du = \mathrm{e}^{ax}, u = \frac{1}{a}\mathrm{e}^{ax}\\ v = \cos (bx), dv = -b\sin (bx) $$ we have $$ \int \mathrm{e}^{ax}\cos (bx)dx = \frac{1}{a}\mathrm{e}^{ax}\cos (bx) + \frac{b}{a}\int \mathrm{e}^{ax}\sin (bx)dx = \frac{1}{a}\mathrm{e}^{ax}\cos (bx) + \frac{b}{a}I $$ so putting it all together we find $$ I = \frac{1}{a}\mathrm{e}^{ax}\sin (bx)-\frac{b}{a}\left[\frac{1}{a}\mathrm{e}^{ax}\cos (bx) + \frac{b}{a}I\right] = \frac{1}{a}\mathrm{e}^{ax}\sin (bx)-\frac{b}{a^2}\mathrm{e}^{ax}\cos (bx) -\frac{b}{a^2}I $$ solve for $I$ i.e. your integral.
Chinny84
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