What is \int_{-\pi}^{\pi}g(t)\,dt equal to?
Direction: Consider the following for the two (02) items that follow : Let f(t)=\ln(t+\sqrt{1+t^{2}}) and g(t)=\tan(f(t)).
- A. -1
- B. 0 ✓
- C. 1/2
- D. 1
Correct Answer: B. 0
Explanation
As established, g(t) is an odd function. By the property of definite integrals, the integral of any odd function over a symmetric interval [-a, a] is always 0. Therefore, \int_{-\pi}^{\pi} g(t)\,dt = 0.
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