How to integrate
$$\int \cos(t)e^{it}dt $$
I tried integrating by parts twice, but it doesn't work because an i shows up in the end and one gets $0=….$
How to integrate
$$\int \cos(t)e^{it}dt $$
I tried integrating by parts twice, but it doesn't work because an i shows up in the end and one gets $0=….$
Making this an answer. Write $\cos(t)=\frac{e^{it}+e^{-it}}2$ and it should be easy.
One way would begin by changing $\cos t$ to $\dfrac{e^{it}+e^{-it}}{2}$. When you multiply that by $e^{it}$ you get $\dfrac{e^{2it}+1}{2}$.
Let's try integrating by parts: $$ \int (\cos t) e^{it} \; dt = \int u\;dv = uv - \int v\;du $$ where $u=\cos t$ and $dv=e^{it}\;dt$, so $du=-\sin t\;dt$ and $v=e^{it}/i= -ie^{it}$. Then $$ uv - \int v\;du = -i(\cos t) e^{it} - \int (\sin t) e^{it}\;dt = -i(\cos t) e^{it} - \int w\;dv $$ some details above may need work. To be continued....with $w= \sin t$ and $v$ as above. This becomes $$ -i(\cos t) e^{it} - \left( wv - \int v\; dw \right) = -i(\cos t) e^{it} - \left( -(\sin t)e^{it} - \int -(\cos t)e^{it} \;dt \right) $$ $$ =-i(\cos t) e^{it} + (\sin t)e^{it} - \int (\cos t)e^{it} \;dt. $$
So we have $$ I = -i(\cos t) e^{it} + (\sin t)e^{it} - I. $$ Solving that for $I$ can probably be left as an excercise. (Remember that when you move $I$ to the other side, "${}+C$" will appear on the right side.)