CUET UG 2025 — Mathematics Calculus

Given integral:

$$\int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$$∫−ππ​esinx+e−sinxesinx​dx

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Let $\displaystyle I = \int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$I=∫−ππ​esinx+e−sinxesinx​dx ... (1)

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Using the property $\displaystyle \int_{-a}^{a} f(x)dx = \int_{-a}^{a} f(-x)dx$∫−aa​f(x)dx=∫−aa​f(−x)dx, replace $x$x with $-x$−x:

$$I = \int_{-\pi}^{\pi} \frac{e^{\sin(-x)}}{e^{\sin(-x)} + e^{-\sin(-x)}}dx$$I=∫−ππ​esin(−x)+e−sin(−x)esin(−x)​dx

Since $\sin(-x) = -\sin(x)$sin(−x)=−sin(x):

$\displaystyle I = \int_{-\pi}^{\pi} \frac{e^{-\sin x}}{e^{-\sin x} + e^{\sin x}}dx$I=∫−ππ​e−sinx+esinxe−sinx​dx ... (2)

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Adding equations (1) and (2):

$$I + I = \int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx + \int_{-\pi}^{\pi} \frac{e^{-\sin x}}{e^{-\sin x} + e^{\sin x}}dx$$I+I=∫−ππ​esinx+e−sinxesinx​dx+∫−ππ​e−sinx+esinxe−sinx​dx

$$2I = \int_{-\pi}^{\pi} \left[\frac{e^{\sin x} + e^{-\sin x}}{e^{\sin x} + e^{-\sin x}}\right]dx$$2I=∫−ππ​[esinx+e−sinxesinx+e−sinx​]dx

$$2I = \int_{-\pi}^{\pi} 1 \, dx$$2I=∫−ππ​1dx

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$$2I = [x]_{-\pi}^{\pi}$$2I=[x]−ππ​

$$2I = \pi - (-\pi)$$2I=π−(−π)

$$2I = 2\pi$$2I=2π

$$I = \pi$$I=π

Therefore, the integral equals $\pi$π.