Q.Integrate the following function: $e^x (\sin x + \cos x)$

The key idea is that the derivative of $e^x \sin x$ is $e^x (\sin x + \cos x)$, which is exactly the integrand.

Step 1: Recall the product rule:

$\frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$. …

Why This Works

When you see an integral like $e^x (\sin x + \cos x)$, your first instinct might be to try integration by parts — and that would work, but it’s unnecessarily long. The trick is to notice a pattern: the derivative of $e^x \sin x$ is $e^x \sin x + e^x \cos x$, which is exactly $e^x (\sin x + \cos x)$. This is a direct consequence of the product rule:

So the integrand is already a derivative. That means the integral is just the original function, plus the constant of integration.

Step-by-Step Solution

1. Observe the structure.

   The integrand is $e^x (\sin x + \cos x)$. Notice that $\sin x$ and $\cos x$ are related by differentiation: $\frac{d}{dx}(\sin x) = \cos x$. So the expression inside the parentheses is $f(x) + f'(x)$ where $f(x) = \sin x$.

2. Recall the product rule for $e^x f(x)$.

   For any differentiable function $f(x)$,

3. Match the pattern. …

Method: The $e^{x}\big(f(x)+f'(x)\big)$ shortcut

Whenever an integrand is $e^{x}$ times a bracket that is a function plus its own derivative, the answer is simply $e^{x}$ times that function:

Steps

Step 1: Isolate the $e^{x}$ factor and look at what multiplies it.

Step 2: Try to split the multiplier as $f(x)+f'(x)$.

Guess a candidate $f(x)$ (often the "nicer" of the two pieces, or the part left after a $\tfrac1x$-type term), then check that its derivative supplies the remaining piece. …

Common Mistakes

Mistake 1: Reaching for by parts and looping forever.

Why it's wrong: $\int e^{x}\sin x\,dx$ by parts is cyclic and slow; here $\sin x+\cos x$ is exactly $f+f'$ with $f=\sin x$. Correct approach: use the $e^{x}(f+f')$ shortcut to write $e^{x}\sin x+C$ at once.

Mistake 2: Choosing $f(x)=\cos x$. …