Evaluate ∫₀^π xsinx dx using integration by parts

Question

Evaluate $\displaystyle\int_0^{\pi} x\sin x \, dx$ using integration by parts.

Solution — Step by Step

Integration by parts formula:

\int u \, dv = uv - \int v \, du

Using the ILATE rule (Inverse trig → Logarithmic → Algebraic → Trigonometric → Exponential), we choose the function that comes first in ILATE as $u$ .

Here we have $x$ (Algebraic) and $\sin x$ (Trigonometric). Algebraic comes before Trigonometric in ILATE.

So: $u = x$ , $dv = \sin x \, dx$

u = x \implies du = dx

dv = \sin x \, dx \implies v = \int \sin x \, dx = -\cos x

\int x\sin x \, dx = uv - \int v \, du

= x(-\cos x) - \int (-\cos x) \, dx

= -x\cos x + \int \cos x \, dx

= -x\cos x + \sin x + C

Now evaluate from $0$ to $\pi$ :

\left[-x\cos x + \sin x\right]_0^{\pi}

At $x = \pi$ :

-\pi\cos\pi + \sin\pi = -\pi(-1) + 0 = \pi

At $x = 0$ :

-0 \cdot \cos 0 + \sin 0 = 0 + 0 = 0

\therefore \int_0^{\pi} x\sin x \, dx = \pi - 0 = \boxed{\pi}

Why This Works

Integration by parts is derived from the product rule for differentiation:

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

Rearranging and integrating both sides:

\int u\frac{dv}{dx} dx = uv - \int v\frac{du}{dx} dx

The key idea is that we “transfer” the differentiation burden from one factor to another. By choosing $u = x$ (which simplifies to a constant when differentiated) and $dv = \sin x \, dx$ (easy to integrate), we reduce the problem from “integrate $x \sin x$ ” to “integrate $\cos x$ ” — much simpler.

Alternative Method — King’s Property

For this particular integral, there’s an elegant shortcut using the King’s property of definite integrals:

\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx

Let $I = \int_0^{\pi} x\sin x \, dx$

Apply King’s property with $a = \pi$ , replacing $x$ with $(\pi - x)$ :

I = \int_0^{\pi} (\pi - x)\sin(\pi - x) \, dx = \int_0^{\pi} (\pi - x)\sin x \, dx

(since $\sin(\pi - x) = \sin x$ )

Adding both expressions:

2I = \int_0^{\pi} x\sin x \, dx + \int_0^{\pi} (\pi - x)\sin x \, dx = \int_0^{\pi} \pi\sin x \, dx

2I = \pi \left[-\cos x\right]_0^{\pi} = \pi(-\cos\pi + \cos 0) = \pi(1 + 1) = 2\pi

I = \pi

Same answer — but this method avoids integration by parts entirely.

In JEE Main, King’s property (sometimes called the “mirror property”) is one of the most time-saving tricks for symmetric definite integrals. Whenever you see $\int_0^{\pi}$ with expressions involving $x$ and $\sin x$ or $\cos x$ , check if King’s property simplifies it before reaching for integration by parts.

Common Mistake

The most common error: forgetting the negative sign when integrating $\sin x$ . Students write $\int \sin x \, dx = \cos x$ instead of $-\cos x$ . This flips the sign of the entire solution. Always double-check: $\frac{d}{dx}(-\cos x) = \sin x$ — yes, the negative sign is correct.

Question

Solution — Step by Step

Why This Works

Alternative Method — King’s Property

Common Mistake

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