Evaluate $\int\limits_0^\frac{\pi}{2}$ sin xdx

Analysis

The question asks us to evaluate a definite integral of the trigonometric function $\sin x$ within the limits of $0$ to $\frac{\pi}{2}$.

Concept

To solve this, we use the fundamental theorem of calculus. The integral of $\sin x$ with respect to $x$ is $-\cos x$. The formula for a definite integral is:

Where $F(x)$ is the antiderivative of $f(x)$.

Step-by-Step Solution

1. Find the antiderivative: The integral of $\sin x$ is $-\cos x$. $$\int \sin x \, dx = -\cos x$$

2. Apply the limits: We evaluate $[-\cos x]$ from $0$ to $\frac{\pi}{2}$: $$\left[ -\cos x \right]_0^{\frac{\pi}{2}}$$

3. Substitute the upper and lower limits: Substitute $\frac{\pi}{2}$ (upper limit) and $0$ (lower limit) into the expression: $$= (-\cos(\frac{\pi}{2})) - (-\cos(0))$$

4. Calculate the trigonometric values: Recall that $\cos(\frac{\pi}{2}) = 0$ and $\cos(0) = 1$. $$= (-0) - (-1)$$ $$= 0 + 1$$ $$= 1$$

Conclusion

The result of the integration is $1$. Comparing this to the given options, it matches option C.

Exam Tip

Always remember your trigonometric values for $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2},$ and $2\pi$. A common mistake students make is forgetting the negative sign when integrating $\sin x$ (remember: the derivative of $\cos x$ is $-\sin x$, so the integral of $\sin x$ must be $-\cos x$).