JEE Mains · Maths · STD 12 - 7.2 definite integral
If the value of the integral \(\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x\) is \(\frac{2}{\pi}\). Then, a value of \(\alpha\) is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^x} d x\) \( I=\int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} d x \) \( \left(\text { using } \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right)\) Add (\(1\)) and (\(II\))…
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