Q.Integrate the following with respect to $x$:

Part (i): $e^{ax}\cos bx$. This is the general formula itself: $\displaystyle\int e^{ax}\cos bx\,dx=\dfrac{e^{ax}}{a^2+b^2}\left[a\cos bx+b\sin bx\right]+c$.

Part (ii): $e^{2x}\sin x$. Here $a=2,\,b=1$, so using $\displaystyle\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}}{a^2+b^2}[a\sin bx-b\cos bx]+c$:

$\displaystyle\int e^{2x}\sin x\,dx=\dfrac{e^{2x}}{4+1}\left[2\sin x-\cos x\right]+c=\dfrac{e^{2x}}5(2\sin x-\cos x)+c$.

Check: $\dfrac d{dx}\left[\dfrac{e^{2x}}5(2\sin x-\cos x)\right]=\dfrac{2e^{2x}}5(2\sin x-\cos x)+\dfrac{e^{2x}}5(2\cos x+\sin x)=\dfrac{e^{2x}}5\left[(4\sin x-2\cos x)+(2\cos x+\sin x)\right]=\dfrac{e^{2x}}5(5\sin x)=e^{2x}\sin x$ ✓.

Part (iii): $e^{-x}\cos2x$. Here $a=-1,\,b=2$: $\displaystyle\int e^{-x}\cos2x\,dx=\dfrac{e^{-x}}{1+4}\left[-\cos2x+2\sin2x\right]+c=\dfrac{e^{-x}}5(2\sin2x-\cos2x)+c$.

Check: $\dfrac d{dx}\left[\dfrac{e^{-x}}5(2\sin2x-\cos2x)\right]=-\dfrac{e^{-x}}5(2\sin2x-\cos2x)+\dfrac{e^{-x}}5(4\cos2x+2\sin2x)=\dfrac{e^{-x}}5\left[(-2\sin2x+\cos2x)+(4\cos2x+2\sin2x)\right]=\dfrac{e^{-x}}5(5\cos2x)=e^{-x}\cos2x$ ✓.

• Using the sine-formula's sign pattern for a cosine integrand or vice versa
• Mixing up $a$ and $b$ when a negative exponent (like $e^{-x}$) is involved