$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}-\int a {\ⅇ}^{a x}\frac{\sin \left(b x\right)}{b} \mathit{\ⅆ}x$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}-\frac{a}{b}\int {\ⅇ}^{a x}\sin \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}-\frac{a}{b}\left({\ⅇ}^{a x}\left(-\frac{\cos \left(b x\right)}{b}\right)-\int a {\ⅇ}^{a x}\left(-\frac{\cos \left(b x\right)}{b}\right) \mathit{\ⅆ}x\right)$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}-\frac{a}{b}\left(-\frac{{\ⅇ}^{a x}\cos \left(b x\right)}{b}\+\frac{a}{b}\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x\right)$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}\+\frac{a}{b^2}{\ⅇ}^{a x}\cos \left(b x\right)-\frac{a^2}{b^2}\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\left(1\+\frac{a^2}{b^2}\right)$ $\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}\+\frac{a}{b^2}{\ⅇ}^{a x}\cos \left(b x\right)$

$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{b^2}{a^2\+b^2}\left(\frac{{\ⅇ}^{a x}\sin \left(b x\right)}{b}\+\frac{a}{b^2}{\ⅇ}^{a x}\cos \left(b x\right)\right)$

$\=\frac{{\ⅇ}^{a x}}{a^2\+b^2}\left(a \cos \left(b x\right)\+b \sin \left(b x\right)\right)$

$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{{\ⅇ}^{a x}\cos \left(b x\right)}{a}-\int \frac{{\ⅇ}^{a x}}{a}\left(-b \sin \left(b x\right)\right) \ⅆx$

$\=\frac{{\ⅇ}^{a x}\cos \left(b x\right)}{a}\+\frac{b}{a}\int {\ⅇ}^{a x}\sin \left(b x\right) \ⅆx$

$\=\frac{{\ⅇ}^{a x}\cos \left(b x\right)}{a}\+\frac{b}{a}\left(\frac{{\ⅇ}^{a x}}{a}\sin \left(b x\right)-\int \frac{{\ⅇ}^{a x}}{a}\left(b \cos \left(b x\right)\right) \ⅆx\right)$

$\=\frac{{\ⅇ}^{a x}\cos \left(b x\right)}{a}\+\frac{b}{a^2}{\ⅇ}^{a x}\sin \left(b x\right)-\frac{b^2}{a^2}\int {\ⅇ}^{a x}\cos \left(b x\right) \ⅆx$

$\left(1\+\frac{b^2}{a^2}\right)$$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\={\ⅇ}^{a x}\left(\frac{\cos \left(b x\right)}{a}\+\frac{b}{a^2}\sin \left(b x\right)\right)$

$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$

$\=\frac{a^2}{a^2\+b^2}{\ⅇ}^{a x}\left(\frac{\cos \left(b x\right)}{a}\+\frac{b}{a^2}\sin \left(b x\right)\right)$

$\=\frac{{\ⅇ}^{a x}}{a^2\+b^2}\left(a \cos \left(b x\right)\+b \sin \left(b x\right)\right)$

Loading Student:-Calculus1

$\int {\ⅇ}^{a x} \cos \left(b x\right) \ⅆx$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int {\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\\ \text{▫}& & \text{1. Apply integration by Parts}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{Parts}}\text{rule}\\ & & \int u\ⅆv=vu-\left(\int v\ⅆu\right)\\ & \text{◦}& \text{First part}\\ & & u=\cos \left(bx\right)\\ & \text{◦}& \text{Second part}\\ & & \mathrm{dv}={\ⅇ}^{ax}\\ & \text{◦}& \text{Differentiate first part}\\ & & \mathrm{du}=\frac{\ⅆ}{\ⅆx}\cos \left(bx\right)\\ & & \mathrm{du}=-\sin \left(bx\right)b\\ & \text{◦}& \text{Integrate second part}\\ & & v=\int {\ⅇ}^{ax}\ⅆx\\ & & v=\frac{{\ⅇ}^{ax}}{a}\\ & & \int {\ⅇ}^{ax}\cos \left(bx\right)\ⅆx=\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}-\left(\int -\frac{{\ⅇ}^{ax}\sin \left(bx\right)b}{a}\ⅆx\right)\\ & & \text{This gives:}\\ & & \frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}-\left(\int -\frac{{\ⅇ}^{ax}\sin \left(bx\right)b}{a}\ⅆx\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{{\ⅇ}^{ax}\sin \left(bx\right)b}{a}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{{\ⅇ}^{ax}\sin \left(bx\right)b}{a}\ⅆx=-\frac{b\left(\int {\ⅇ}^{ax}\sin \left(bx\right)\ⅆx\right)}{a}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}+\frac{b\left(\int {\ⅇ}^{ax}\sin \left(bx\right)\ⅆx\right)}{a}\\ \text{▫}& & \text{3. Apply integration by Parts}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{Parts}}\text{rule}\\ & & \int u\ⅆv=vu-\left(\int v\ⅆu\right)\\ & \text{◦}& \text{First part}\\ & & u=\sin \left(bx\right)\\ & \text{◦}& \text{Second part}\\ & & \mathrm{dv}={\ⅇ}^{ax}\\ & \text{◦}& \text{Differentiate first part}\\ & & \mathrm{du}=\frac{\ⅆ}{\ⅆx}\sin \left(bx\right)\\ & & \mathrm{du}=b\cos \left(bx\right)\\ & \text{◦}& \text{Integrate second part}\\ & & v=\int {\ⅇ}^{ax}\ⅆx\\ & & v=\frac{{\ⅇ}^{ax}}{a}\\ & & \int {\ⅇ}^{ax}\sin \left(bx\right)\ⅆx=\frac{\sin \left(bx\right){\ⅇ}^{ax}}{a}-\left(\int \frac{{\ⅇ}^{ax}b\cos \left(bx\right)}{a}\ⅆx\right)\\ & & \text{This gives:}\\ & & \frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}+\frac{b\left(\frac{\sin \left(bx\right){\ⅇ}^{ax}}{a}-\left(\int \frac{{\ⅇ}^{ax}b\cos \left(bx\right)}{a}\ⅆx\right)\right)}{a}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{{\ⅇ}^{ax}b\cos \left(bx\right)}{a}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{{\ⅇ}^{ax}b\cos \left(bx\right)}{a}\ⅆx=\frac{b\left(\int {\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\right)}{a}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}+\frac{b\left(\frac{\sin \left(bx\right){\ⅇ}^{ax}}{a}-\frac{b\left(\int {\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\right)}{a}\right)}{a}\\ \text{▫}& & \text{5. Solve the equation algebraically}\\ & \text{◦}& \text{Define}\\ & & F=\int {\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\\ & \text{◦}& \text{Solve for}F\\ & & F=\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}+\frac{b\left(\frac{\sin \left(bx\right){\ⅇ}^{ax}}{a}-\frac{bF}{a}\right)}{a}\\ & & \text{This gives:}\\ & & \frac{{\ⅇ}^{ax}\left(\cos \left(bx\right)a+\sin \left(bx\right)b\right)}{a^2+b^2}\end{array}$

Table 6.1.4(a)   Annotated stepwise solution in a specific case

Initialize

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\:$$$

$\int {\ⅇ}^{a x}\cos \left(b x\right) \mathit{\ⅆ}x$$\stackrel{\text{assign to a name}}{\to }$$Q$

First parts integration

$q_1≔\mathrm{Parts}\left(Q\,{\ⅇ}^{a x}\right)$

$\frac{\sin \left(bx\right){\ⅇ}^{ax}}{b}-\left(\∫\frac{\sin \left(bx\right)a{\ⅇ}^{ax}}{b}\ⅆx\right)$

Second parts integration

$q_2≔\mathrm{Parts}\left(q_1\,{\ⅇ}^{a x}\right)$

$\frac{\sin \left(bx\right){\ⅇ}^{ax}}{b}\+\frac{a\cos \left(bx\right){\ⅇ}^{ax}}{b^2}\+\∫\left(-\frac{a^2\cos \left(bx\right){\ⅇ}^{ax}}{b^2}\right)\ⅆx$

Form the equation $Q-F\left(Q\right)\=0$ 

$q_3≔Q-q_2\=0$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx-\frac{\sin \left(bx\right){\ⅇ}^{ax}}{b}-\frac{a\cos \left(bx\right){\ⅇ}^{ax}}{b^2}-\left(\∫\left(-\frac{a^2\cos \left(bx\right){\ⅇ}^{ax}}{b^2}\right)\ⅆx\right)\=0$

Simplify the left-hand side

$q_4≔\mathrm{expand}\left(q_3\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx-\frac{\sin \left(bx\right){\ⅇ}^{ax}}{b}-\frac{a\cos \left(bx\right){\ⅇ}^{ax}}{b^2}\+\frac{a^2\left(\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\right)}{b^2}\=0$

Isolate the integral $Q$ 

$q_5≔\mathrm{isolate}\left(q_4\,Q\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\=\frac{\frac{\sin \left(bx\right){\ⅇ}^{ax}}{b}\+\frac{a\cos \left(bx\right){\ⅇ}^{ax}}{b^2}}{1\+\frac{a^2}{b^2}}$

Simplify the right-hand side

$\mathrm{normal}\left(q_5\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\=\frac{{\ⅇ}^{ax}\left(\cos \left(bx\right)a\+\sin \left(bx\right)b\right)}{a^2\+b^2}$

Table 6.1.4(b)   Stepwise solution in the general case, with $u\={\ⅇ}^{a x}$

First parts integration

$q_6≔\mathrm{Parts}\left(Q\,\cos \left(b x\right)\right)$

$\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}-\left(\∫\left(-\frac{{\ⅇ}^{ax}\sin \left(bx\right)b}{a}\right)\ⅆx\right)$

Second parts integration

$q_7≔\mathrm{Parts}\left(q_6\,\sin \left(b x\right)\right)$

$\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}\+\frac{{\ⅇ}^{ax}b\sin \left(bx\right)}{a^2}\+\∫\left(-\frac{{\ⅇ}^{ax}{b}^2\cos \left(bx\right)}{a^2}\right)\ⅆx$

Form the equation $Q-G\left(Q\right)\=0$ 

$q_8≔Q-q_7\=0$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx-\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}-\frac{{\ⅇ}^{ax}b\sin \left(bx\right)}{a^2}-\left(\∫\left(-\frac{{\ⅇ}^{ax}{b}^2\cos \left(bx\right)}{a^2}\right)\ⅆx\right)\=0$

Simplify the left-hand side

$q_9≔\mathrm{expand}\left(q_8\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx-\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}-\frac{{\ⅇ}^{ax}b\sin \left(bx\right)}{a^2}\+\frac{b^2\left(\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\right)}{a^2}\=0$

Isolate the integral $Q$ 

$q_{10}≔\mathrm{isolate}\left(q_9\,Q\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\=\frac{\frac{{\ⅇ}^{ax}\cos \left(bx\right)}{a}\+\frac{{\ⅇ}^{ax}b\sin \left(bx\right)}{a^2}}{1\+\frac{b^2}{a^2}}$

Simplify the right-hand side

$\mathrm{normal}\left(q_{10}\right)$

$\∫{\ⅇ}^{ax}\cos \left(bx\right)\ⅆx\=\frac{{\ⅇ}^{ax}\left(\cos \left(bx\right)a\+\sin \left(bx\right)b\right)}{a^2\+b^2}$

Table 6.1.4(c)   Stepwise solution in the general case, with $u$ taken as the trig term in each integration