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GUJCET · Maths · Integrals
\(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x=\) __________.
- A \(\frac{\pi}{4}+\frac{1}{2}\)
- B \(\frac{\pi}{4}-\frac{1}{2}\)
- C \(\frac{\pi}{4}-1\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1-\cos(2x)}{2} d x \) \( = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \)
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