Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$ \int e^{ax}\cos(bx)\,\mathrm dx = \frac1{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,\mathrm dx$$

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

Where does the $1$ in $\displaystyle \left(1 + \frac{b^2}{a^2}\right)$ come from?

share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

Yeah so they have taken the quantity $-b^{2}/a^{2}$ to the Left hand side. So you have the LHS as $$1 \cdot \int e^{ax}\cos(bx) \rm{dx} + \frac{b^{2}}{a^{2}} \int e^{ax}\cos(bx) \ \text{dx} = \Bigl(1+\frac{b^{2}}{a^{2}}\Bigr)\int e^{ax}\cos(bx) \ \text{dx}$$

share|cite|improve this answer
If $a \neq 0$ and you multiply $a$ with $1$ you get $a$ itself. – user9413 May 16 '11 at 8:53
I guess there is a 1 there. Thank you =) – Louis May 16 '11 at 9:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.