Area between curves y = sin(x) and y = cos(x) from 0 to π/2

Question

Find the area enclosed between the curves $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \pi/2$ .

(JEE Main 2023)

Solution — Step by Step

Set $\sin x = \cos x$ :

\tan x = 1 \implies x = \frac{\pi}{4}

So the curves cross at $x = \pi/4$ in the interval $[0, \pi/2]$ .

On $[0, \pi/4]$ : $\cos x > \sin x$ (e.g., at $x = 0$ : $\cos 0 = 1 > 0 = \sin 0$ )

On $[\pi/4, \pi/2]$ : $\sin x > \cos x$ (e.g., at $x = \pi/2$ : $\sin(\pi/2) = 1 > 0 = \cos(\pi/2)$ )

\text{Area} = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx

First integral:

\int_0^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_0^{\pi/4}

= \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1) = \sqrt{2} - 1

Second integral:

\int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}

= (0 - 1) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -1 + \sqrt{2} = \sqrt{2} - 1

Total area:

\text{Area} = (\sqrt{2} - 1) + (\sqrt{2} - 1) = \boxed{2(\sqrt{2} - 1)} \approx 0.828

Why This Works

The area between two curves is always $\int |f(x) - g(x)| \, dx$ — we need the absolute value because area is always positive. The practical way to handle the absolute value is to split the interval at the intersection points and determine which function is larger in each subinterval.

Notice the beautiful symmetry here: both subregions have equal area ( $\sqrt{2} - 1$ each). This happens because $\sin x$ and $\cos x$ are the same function shifted by $\pi/2$ , and our region is symmetric about $x = \pi/4$ .

Alternative Method — Single integral with absolute value

\text{Area} = \int_0^{\pi/2} |\cos x - \sin x| \, dx

Since we know the sign change occurs at $x = \pi/4$ :

= \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx

This is the same computation, just written more compactly.

For JEE, the answer $2(\sqrt{2} - 1)$ is a commonly appearing numerical value in area problems involving trig functions. Also remember: the area between $\sin x$ and $\cos x$ over one full “crossing” (from one intersection to the next) is always $2(\sqrt{2} - 1)$ . This can save time in MCQs.

Common Mistake

The most frequent error: students compute $\int_0^{\pi/2} (\cos x - \sin x) \, dx$ without splitting at the intersection point. This gives $[\sin x + \cos x]_0^{\pi/2} = (1 + 0) - (0 + 1) = 0$ . The zero result should immediately raise a red flag — the area between two curves that cross cannot be zero. The positive and negative parts cancel out. Always split the integral at intersection points.

Question

Solution — Step by Step

Why This Works

Alternative Method — Single integral with absolute value

Common Mistake

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