Definite integral of sin x between 0 and π with geometric interpretation.

Evaluate ∫sin x dx from 0 to π

Question

Evaluate the definite integral:

\int_0^{\pi} \sin x \, dx

This appeared in CBSE 2024 Board Exam and is a standard 2-mark question. Clean setup, clean answer — but a surprisingly common place to lose marks.

Solution — Step by Step

The antiderivative of $\sin x$ is $-\cos x$ . We can verify this by differentiating: $\frac{d}{dx}(-\cos x) = \sin x$ . Always confirm your antiderivative before applying limits.

\int_0^{\pi} \sin x \, dx = \Big[-\cos x\Big]_0^{\pi}

Substitute $x = \pi$ into $-\cos x$ :

-\cos(\pi) = -(-1) = +1

Remember: $\cos \pi = -1$ , so the two negatives cancel to give $+1$ .

Substitute $x = 0$ into $-\cos x$ :

-\cos(0) = -(1) = -1

$\cos 0 = 1$ , so we get $-1$ here.

\Big[-\cos x\Big]_0^{\pi} = (-\cos \pi) - (-\cos 0) = (1) - (-1) = 2

The final answer is 2.

Why This Works

The integral $\int_0^{\pi} \sin x \, dx$ is asking for the net signed area between $y = \sin x$ and the x-axis from $0$ to $\pi$ .

On the interval $[0, \pi]$ , $\sin x \geq 0$ throughout — the curve sits entirely above the x-axis. So the “net” area is just the actual area, with no cancellation. The answer 2 represents exactly one “hump” of the sine wave.

This is why the answer is a positive whole number — it’s geometrically satisfying. If you had integrated from $0$ to $2\pi$ instead, the positive hump ( $0$ to $\pi$ ) and the negative hump ( $\pi$ to $2\pi$ ) would cancel, giving zero.

The area under one complete arch of $\sin x$ is always 2. This is a fixed result worth memorising — it appears in JEE Main MCQs where they ask for area using integration, and knowing this saves calculation time.

Alternative Method — Using Geometry / Symmetry Check

We can cross-verify using the property of definite integrals. Since $\sin(\pi - x) = \sin x$ , the function is symmetric about $x = \pi/2$ on $[0, \pi]$ .

This means the area from $0$ to $\pi/2$ equals the area from $\pi/2$ to $\pi$ . We can compute just half:

\int_0^{\pi/2} \sin x \, dx = \Big[-\cos x\Big]_0^{\pi/2} = (-\cos \tfrac{\pi}{2}) - (-\cos 0) = 0 - (-1) = 1

Double it: $2 \times 1 = 2$ . Same answer, confirmed.

This symmetry check is useful in MCQ papers when you want to verify quickly without redoing the full calculation.

Common Mistake

The most frequent error: students write $\int_0^{\pi} \sin x \, dx = [\cos x]_0^{\pi}$ — they forget the negative sign in the antiderivative.

This gives $\cos \pi - \cos 0 = -1 - 1 = -2$ , which is wrong in sign. The antiderivative of $\sin x$ is $-\cos x$ , not $+\cos x$ . The negative is not optional — it comes directly from the chain rule in reverse. If you’re unsure, differentiate $-\cos x$ on your rough sheet before proceeding.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Geometry / Symmetry Check

Common Mistake

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