The image you uploaded contains a definite integral. The expression looks like:

$\int_{0}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$

Step-by-Step Solution:

1. Split the Integral: We can split the integral into two separate integrals: $\int_{0}^{\frac{\pi}{2}} \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \cos x \, dx$

2. Evaluate Each Integral:

   • For the first integral: $\int \sin x \, dx = -\cos x + C$ So, $\int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}}$ Substituting the limits: $-\cos\left(\frac{\pi}{2}\right) + \cos(0) = -0 + 1 = 1$

   • For the second integral: $\int \cos x \, dx = \sin x + C$ So, $\int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}}$ Substituting the limits: $\sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$

3. Combine the Results: Now subtract the second result from the first: $1 - 1 = 0$

Final Answer:

$\int_{0}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx = 0$

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Follow-Up Questions:

1. What are the basic properties of definite integrals?
2. How can you evaluate an integral with more complex trigonometric functions?
3. What is the significance of splitting integrals in solving them?
4. How do you determine the antiderivative of trigonometric functions?
5. What are some common mistakes to avoid when solving definite integrals?

Tip:

Always double-check the limits of integration when solving definite integrals, as errors in limits can lead to incorrect results.