The value of _0^e^ x x(1+^2x)(e^ x+e^- x),dx is equal to:

Question

The value of $\displaystyle\int_0^\pi\frac{e^{\cos x}\sin x}{(1+\cos^2x)(e^{\cos x}+e^{-\cos x})}\,dx$ is equal to:

Accepted Answer

By King's rule ($x\to\pi-x$, so $\cos x\to-\cos x$, $\sin x\to\sin x$) and adding both forms:

$$2I = \int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$$

With $u=\cos x$:

$$2I = \int_{-1}^1\frac{du}{1+u^2} = \big[\tan^{-1}u\big]_{-1}^1 = \frac\pi2$$

$$I = \frac\pi4$$

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