JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of the integral \(\int \limits_{-1}^{1} \log _{ e }(\sqrt{1- x }+\sqrt{1+ x }) dx\) is equal to:
- A \(2 \log _{e} 2+\frac{\pi}{4}-1\)
- B \(\frac{1}{2} \log _{e} 2+\frac{\pi}{4}-\frac{3}{2}\)
- C \(2 \log _{\mathrm{e}} 2+\frac{\pi}{2}-\frac{1}{2}\)
- D \(\log _{e} 2+\frac{\pi}{2}-1\)
Answer & Solution
Correct Answer
(D) \(\log _{e} 2+\frac{\pi}{2}-1\)
Step-by-step Solution
Detailed explanation
\(\text { Let } I=2 \int_{0}^{1} \underbrace{\operatorname{In}(\sqrt{1-x}+\sqrt{1+x})}_{(I)} \underbrace{1}_{(II)} d x\) \(\quad\quad\quad\quad\quad\quad\quad\quad(I.B.P.)\) \(\therefore I=\mid x \cdot I n(\sqrt{1-x}+\sqrt{1-x})_{0}^{1}\)…
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