_0^() (e^( x))/(e^( x) + e^(- x)) dx = Option 1: /4 Option 2: /3 Option 3: /6 Option 4: /2

Answer: /2. Solution: Let I = _0^() (e^( x))/(e^( x) + e^(- x))dx. Using _0^a f(x)dx = _0^a f(a-x)dx and (-x) = - x: I = _0^() (e^(- x))/(e^(- x) + e^( x))dx. Adding both: 2I = _0^() 1\ dx = . So I = /2.

Calculus

Integrals

Medium

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx =$

$\pi/4$

$\pi/3$

$\pi/6$

$\pi/2$

✅ Correct Option: 4

Let $I = \int_0^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}}dx$ .

Using $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ and $\cos(\pi-x) = -\cos x$ :

$I = \int_0^{\pi} \frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}}dx$ .

Adding both: $2I = \int_0^{\pi} 1\ dx = \pi$ .

So $I = \pi/2$ .

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The value of $\int_{0}^{1} \frac{a-b x^{2}}{\left(a+b x^{2}\right)^{2}} d x$ is :

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The value of the integral $I = \int_0^1 \frac{1}{\sqrt{1+3\sqrt{x}}} dx$ is

∫0πecos⁡xecos⁡x+e−cos⁡xdx=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx =∫0π​ecosx+e−cosxecosx​dx=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx =$