Q.Find $\int e^x \sin x\, dx$

The key idea is to use integration by parts twice, then solve for the original integral algebraically.

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

Step 1: Apply integration by parts with $u = \sin x$, $dv = e^x dx$. Then $du = \cos x \, dx$, $v = e^x$.

Step 2: Apply integration by parts again to $\int e^x \cos x \, dx$, with $u = \cos x$, $dv = e^x dx$. Then $du = -\sin x \, dx$, $v = e^x$.

Step 3: Substitute back into Step 1. …

Why This Approach Works

When you see a product of two different kinds of functions — here, an exponential $e^x$ and a trigonometric $\sin x$ — integration by parts is the natural tool. The formula is:

The trick is that neither $e^x$ nor $\sin x$ simplifies when differentiated or integrated — they just cycle between each other. Differentiating $\sin x$ gives $\cos x$, and integrating $e^x$ gives $e^x$ again. If we apply integration by parts twice, we'll end up back at $\int e^x \sin x \, dx$, which we can then solve for like an algebraic equation.

Step-by-Step Solution

1. First integration by parts

Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$.

Applying the formula:

We now have a new integral $\int e^x \cos x \, dx$ to handle.

2. Second integration by parts

For $\int e^x \cos x \, dx$, let $u = \cos x$ and $dv = e^x \, dx$. Then $du = -\sin x \, dx$ and $v = e^x$.

So:

3. Substitute back

Plug this result into the expression from step 1:

Simplify:

$\int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx$ …

Method: Cyclic Integration by Parts (Exponential × Trigonometric)

Use this when the integrand is $e^{ax}\sin bx$ or $e^{ax}\cos bx$: applying parts twice reproduces the original integral, letting you solve for it algebraically.

Steps

Step 1: Apply by-parts once.

Neither factor simplifies, so pick either as $u$ consistently. For $I=\int e^x\sin x\,dx$, take $u=\sin x$, $dv=e^x dx$; you get $I=e^x\sin x-\int e^x\cos x\,dx$.

Step 2: Apply by-parts again, keeping the same type choice. …

Common Mistakes

Mistake 1: Switching the $u$/$dv$ roles between the two applications.

Why it's wrong: mixing choices makes the integral cycle back to $0=0$ instead of a solvable equation. Correct approach: keep $e^x$ (or the trig factor) as $dv$ both times.

Mistake 2: Forgetting the sign when the original integral reappears. …