TS EAMCET · Maths · Definite Integration
\(\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x\) is equal to :
- A 0
- B 1
- C \(\frac{e^2-1}{2 e}\)
- D \(\frac{e^2+2}{2 e}\)
Answer & Solution
Correct Answer
(C) \(\frac{e^2-1}{2 e}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x\) \(=\int_{-1}^1 \frac{e^x+e^{-x}}{2\left(1+e^{2 x}\right)} d x\left(\because \cosh x=\frac{e^x+e^{-x}}{2}\right)\) \(=\frac{1}{2} \int_{-1}^1 \frac{1+e^{2 x}}{\left(1+e^{2 x}\right) e^x} \cdot d x\)…
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