Set of 12 integration problems for calculus practice.

Let’s solve problem #1:
∫ eˣ sin x dx

This is a classic integration by parts problem — and it’s tricky because you’ll need to do it *twice*, then solve for the original integral.

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Step 1: Recall Integration by Parts Formula

We use:
∫ u dv = uv - ∫ v du

We need to pick u and dv wisely. Let’s try:

Let
→ u = sin x → du = cos x dx
→ dv = eˣ dx → v = eˣ

Then:

∫ eˣ sin x dx = eˣ sin x - ∫ eˣ cos x dx ...(Equation A)

Now we have a new integral: ∫ eˣ cos x dx — which looks similar. We’ll apply integration by parts again.

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Step 2: Apply Integration by Parts Again on ∫ eˣ cos x dx

Let
→ u = cos x → du = -sin x dx
→ dv = eˣ dx → v = eˣ

So:

∫ eˣ cos x dx = eˣ cos x - ∫ eˣ (-sin x) dx
= eˣ cos x + ∫ eˣ sin x dx ...(Equation B)

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Step 3: Plug Equation B into Equation A

From Equation A:

∫ eˣ sin x dx = eˣ sin x - [eˣ cos x + ∫ eˣ sin x dx]

Simplify:

∫ eˣ sin x dx = eˣ sin x - eˣ cos x - ∫ eˣ sin x dx

Now, notice that the same integral appears on both sides! Let’s call I = ∫ eˣ sin x dx

So:

I = eˣ sin x - eˣ cos x - I

Add I to both sides:

2I = eˣ sin x - eˣ cos x

Divide by 2:

I = (eˣ sin x - eˣ cos x)/2 + C

Factor out eˣ:

I = (eˣ / 2)(sin x - cos x) + C

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✔ Final Answer:
(eˣ / 2)(sin x - cos x) + C