Evaluate $\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x$

1. Analyze the question

We are asked to evaluate the definite integral of the function $\sin x$ with respect to $x$, within the limits of $0$ to $\frac{\pi}{2}$.

2. Explain the concept

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

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

3. Show the solution

Step 1: Find the integral. The integral of $\sin x$ is $-\cos x$.

Step 2: Apply the limits $0$ and $\frac{\pi}{2}$.

Step 3: Substitute the upper and lower limits.

Step 4: Evaluate the trigonometric values. Recall that:

• $\cos(\frac{\pi}{2}) = 0$
• $\cos(0) = 1$

Substitute these values into the expression:

4. Identify the answer

The result of the integration is $1$. Comparing this to the given options: A. -2 B. 2 C. 1 D. -1

The correct option is C.

5. Tip

When dealing with trigonometric integrals, always remember the "negative sign" rule: the derivative of $\cos x$ is $-\sin x$, so the integral of $\sin x$ must be $-\cos x$. Don't forget to distribute the negative sign when subtracting the lower limit!