Evaluate the integral: I = integral e^axcos²x dx

Evaluate the integral: $$ I = \int e^{ax}\cos^2x\,dx. $$

Step 1 · Apply the power-reduction identity

Using $\cos^2x=\dfrac{1+\cos2x}{2}$:

$$ I = \frac12\int e^{ax}\,dx + \frac12\int e^{ax}\cos2x\,dx. $$

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, 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

Additional Notes

• The integrand is defined for all real $x$ (assuming $a\neq0$).
• This is the mirror image of Problem 171, with only the sign in front of the $e^{ax}\cos2x$ piece flipped (plus here, minus there) — a direct consequence of the plus/minus sign difference between $\cos^2x$ and $\sin^2x$'s power-reduction identities.
• The result can be checked by differentiating the final expression and confirming, via the identity $1+\cos2x=2\cos^2x$, that it returns $e^{ax}\cos^2x$.