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∫₀^a sin x dx = 1 − cos a

Integrals · Class XII

Slide a , the running area under sin x bounces between 0 and 2 because of cancellation.

3.14
type a value
09.42
Live values
  • Integral2
  • Max possible2
xy
  • y = sin x
  • F(x) = 1 − cos x

Formulas in this lab

  • Integral
    ∫0asin⁡x dx=1−cos⁡a\int_0^a \sin x\,dx = 1 - \cos a
  • Max possible
    22
Tip: At a = π the integral hits 2 , the full positive half-cycle area.

Frequently asked questions

▶What is $\int_0^a \sin x\,dx$?

Since the antiderivative of $\sin x$ is $-\cos x$, you get $\int_0^a \sin x\,dx = -\cos a + \cos 0 = 1 - \cos a$. For $a = \pi$, the value is $2$ — exactly the area of one positive half of the sine wave.

▶How do I use the $\int \sin x$ lab?

Slide the upper limit $a$ between $0$ and $2\pi$. The lab shades the area under $y = \sin x$ and computes $1 - \cos a$. Try $a = \pi/2$: integral is $1$, the area of the quarter wave.

▶Why does $\int_0^{2\pi}\sin x\,dx = 0$?

The positive area from $0$ to $\pi$ and the equal negative area from $\pi$ to $2\pi$ cancel. So the signed integral is zero, even though the geometric area is $4$. Students confuse signed integral with absolute area on JEE — read each problem carefully.

▶Where does $\int\sin x\,dx$ appear in real life?

Average power in AC circuits uses $\int_0^T \sin^2(\omega t)\,dt$; the related building block $\int\sin x\,dx$ appears in deriving capacitor charge and pendulum motion energies. Even the area of a circular segment uses sine integrals indirectly.