Q.Integrate the following function: $e^{2x} \sin x$

The key idea is to treat this as a product of two functions and apply integration by parts twice, then solve for the original integral.

Let $I = \int e^{2x} \sin x \, dx$.

Step 1: Choose $u = \sin x$, $dv = e^{2x} dx$. Then $du = \cos x \, dx$, $v = \frac{1}{2} e^{2x}$.

Step 2: Apply integration by parts again to $\int e^{2x} \cos x \, dx$. Let $u = \cos x$, $dv = e^{2x} dx$, so $du = -\sin x \, dx$, $v = \frac{1}{2} e^{2x}$.

Step 3: Substitute back into the expression for $I$:

$I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I \right)$ …

Why This Approach Works

When you see a product of an exponential and a trigonometric function, your first instinct might be to try substitution — but that won't help here because neither function is the derivative of the other in a simple way. The key insight is that integration by parts can reduce the complexity step by step, but because both $e^{2x}$ and $\sin x$ are "cyclic" under differentiation (they loop back to themselves after two derivatives), we end up with the original integral reappearing. That lets us treat it as an algebraic equation and solve for the unknown integral.

Step-by-Step Solution

1. Set up the integral and choose parts.

   Let $I = \int e^{2x} \sin x \, dx$.

   For integration by parts, we need $u$ and $dv$. A good rule: pick $u$ as the function that simplifies when differentiated. Here, both $e^{2x}$ and $\sin x$ are fine, but let's choose:

   $u = \sin x$, $dv = e^{2x} dx$.

   Then $du = \cos x \, dx$, and $v = \int e^{2x} dx = \frac{1}{2} e^{2x}$.

2. Apply integration by parts the first time.

   The formula $\int u \, dv = uv - \int v \, du$ gives:

3. Now we need $\int e^{2x} \cos x \, dx$ — call it $J$. Apply integration by parts again to $J$. This time, let $u = \cos x$, $dv = e^{2x} dx$. Then $du = -\sin x \, dx$, $v = \frac{1}{2} e^{2x}$. So:

4. Notice the original integral $I$ has reappeared. The last term is exactly $\frac{1}{2} I$. So we have:

5. Substitute $J$ back into the expression for $I$. From step 2: $I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} J$. Replace $J$: $I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{2} I \right).$ …

Method: Cyclic integration by parts (exponential times sine/cosine)

For $e^{ax}\sin bx$ or $e^{ax}\cos bx$, two rounds of by parts reproduce the original integral, which you then solve algebraically.

Steps

Step 1: Call the integral $I$ and integrate by parts once.

Let $I=\int e^{2x}\sin x\,dx$. Take $u=\sin x$, $dv=e^{2x}dx$ (either choice works if kept consistent), producing an $e^{2x}\cos x$ integral.

Step 2: Integrate by parts a second time, consistently.

Apply by parts to the new integral with the same type of choice. A term equal to a multiple of the original $I$ reappears.

Step 3: Solve for $I$ algebraically.

You reach a relation like $I=(\text{boundary terms})-k\,I$. Collect the $I$ terms: …

Common Mistakes

Mistake 1: Switching the by-parts choice between the two passes.

Why it's wrong: differentiating the exponential in one pass and integrating it in the next unravels the loop back to $0=0$. Correct approach: keep the same role (always $u=$ trig, or always $u=$ exponential) both times.

Mistake 2: Panicking when the original integral reappears.

Why it's wrong: the reappearance is the whole mechanism, not a dead end. Correct approach: treat $I$ as an unknown and solve the linear equation for it. …