Evaluate the integral: I = integral e^axsin²x dx
Evaluate the integral: $$ I = \int e^{ax}\sin^2x\,dx. $$
Explanation of Method
The power-reduction identity $\sin^2x=\tfrac{1-\cos2x}{2}$ splits the integral into a plain exponential term and an $e^{ax}\cos2x$ term, the latter resolved by Problem 170's general formula with $b=2$.
- Rewrite $\sin^2x$ using the power-reduction identity.
- Split the integral into two pieces.
- Apply the $\int e^{ax}\cos bx\,dx$ formula (Problem 170) with $b=2$.
- Combine the results.
Step 1 · Apply the power-reduction identity
Using $\sin^2x=\dfrac{1-\cos2x}{2}$:
$$ I = \frac12\int e^{ax}\,dx - \frac12\int e^{ax}\cos2x\,dx. $$
Step 2 · Evaluate the first piece
$$ \int e^{ax}\,dx = \frac{1}{a}e^{ax}. $$
Step 3 · Evaluate the second piece
From Problem 170's general formula $\int e^{ax}\cos bx\,dx=\dfrac{1}{a^2+b^2}(ae^{ax}\cos bx+be^{ax}\sin bx)$, with $b=2$:
$$ \int e^{ax}\cos2x\,dx = \frac{1}{a^2+4}\left(ae^{ax}\cos2x + 2e^{ax}\sin2x\right). $$
Step 4 · Combine
$$ I = \frac{1}{2a}e^{ax} - \frac{1}{2(a^2+4)}\left(ae^{ax}\cos2x + 2e^{ax}\sin2x\right) + C. $$
Summary Table
| Piece | Value |
|---|---|
| $\tfrac12\int e^{ax}dx$ | $\tfrac{1}{2a}e^{ax}$ |
| $\tfrac12\int e^{ax}\cos2x\,dx$ | $\tfrac{1}{2(a^2+4)}(ae^{ax}\cos2x+2e^{ax}\sin2x)$ |
Additional Notes
- The integrand is defined for all real $x$ (assuming $a\neq0$).
- Rather than re-deriving the self-referential integration by parts for $\int e^{ax}\cos2x\,dx$ from scratch, this problem reuses the general formula already established in Problem 170 with $b=2$ — a good example of building on previously solved results.
- The result can be checked by differentiating the final expression and confirming, via the identity $1-\cos2x=2\sin^2x$, that it returns $e^{ax}\sin^2x$.
Result
The solution to the integral is: