Question

$\int e^{ax}\cos bx$ is

• A. $\frac{e^{ax}}{a^2+b^2}[a\cos bx+b\sin bx]+c$
• B. $\frac{e^{ax}}{a^2+b^2}[a^2\sin bx-b^2\cos bx]+c$
• C. $\frac{e^{ax}}{a+b}[a\cos bx-b\sin bx]+c$
• D. $\frac{e^{ax}}{a^2+b^2}[a\sin bx-b\cos bx]+c$

Answer 1

The correct option is A $\frac{e^{ax}}{a^2+b^2}[a\cos bx+b\sin bx]+c$

We'll solve the above problem by using integration by parts,

$I=\int e^{ax}\cos bxdx$

$I=e^{ax}\int \cos bxdx-\int a{e}^{ax}\left(\int \cos bxdx\right)dx$

$I=e^{ax}\left[\frac{\sin bx}{b}\right]-\int a{e}^{ax}\left(\frac{\sin bx}{b}\right)dx$

$I=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}\int e^{ax}\sin bxdx$

Let's use integration by parts again,

$I=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}\left[e^{ax}\int \sin bxdx-\int a{e}^{ax}\left(\int \sin bxdx\right)dx\right]$

$I=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}[e^{ax}\left[\frac{-\cos bx}{b}\right]-\int a{e}^{ax}\left(\frac{-\cos bx}{b}\right)dx]$

$I=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}[\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}\int e^{ax}\cos bxdx]$

But we know that,

$I=\int e^{ax}\cos bxdx$

So,

$I=\frac{e^{ax}\sin bx}{b}-\frac{a}{b}[\frac{-e^{ax}\cos bx}{b}+\frac{a}{b}I]I=\frac{e^{ax}\sin bx}{b}+\frac{a{e}^{ax}\cos bx}{b^2}-\frac{a^2}{b^2}I(1+\frac{a^2}{b^2})I=\frac{e^{ax}}{b^2}[a\cos bx+b\sin bx]\therefore I=\frac{e^{ax}}{a^2+b^2}[a\cos bx+b\sin bx]+c$.