KCET · Maths · Area Under Curves
The value of \(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x\) is
- A 2
- B 0
- C 1
- D \(-2\)
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
We have, \(\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x\)
\(=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{\cos (-x)}{\left(1+e^{-x}\right)}\right) d x\)
\(\left[\because \int_{-a}^{a} f(x) d x=\int_{0}^{a}(f(x)+f(-x)) d x\right]\)
\(=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{e^{x} \cos x}{1+e^{x}}\right) d x\)
\(=\int_{0}^{\pi / 2} \cos x\left(\frac{1+e^{x}}{1+e^{x}}\right) d x\)
\(=\int_{0}^{\pi / 2} \cos x d x=(\sin x)_{0}^{\pi / 2}\)
\(=\left(\sin \frac{\pi}{2}-\sin 0\right)=1\)
\(=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{\cos (-x)}{\left(1+e^{-x}\right)}\right) d x\)
\(\left[\because \int_{-a}^{a} f(x) d x=\int_{0}^{a}(f(x)+f(-x)) d x\right]\)
\(=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{e^{x} \cos x}{1+e^{x}}\right) d x\)
\(=\int_{0}^{\pi / 2} \cos x\left(\frac{1+e^{x}}{1+e^{x}}\right) d x\)
\(=\int_{0}^{\pi / 2} \cos x d x=(\sin x)_{0}^{\pi / 2}\)
\(=\left(\sin \frac{\pi}{2}-\sin 0\right)=1\)
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