Mathematics

Evaluate: $$\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^{2}x}dx$$


SOLUTION
$$\begin{array}{l}\text { Let } I=\displaystyle \int_{0}^{\pi} \dfrac{x \sin x}{1+\cos ^{2} x} \mathrm{d} \mathrm{x} \quad \Rightarrow \quad I=\displaystyle \int_{0}^{\pi} \dfrac{(\pi-x) \sin (\pi-x) \mathrm{d} x}{1+\cos ^{2}(\pi-x)} \\=\displaystyle \int_{0}^{\pi} \dfrac{(\pi-x) \sin x \mathrm{d} x}{1+\cos ^{2} x}=\pi \int_{0}^{\pi} \dfrac{\sin x \mathrm{d} x}{1+\cos ^{2} x}-I \\\text { or } \displaystyle 2 I=\pi \int_{0}^{\pi} \dfrac{\sin x \mathrm{dx}}{1+\cos ^{2} x} \text { or } I=\dfrac{\pi}{2} \int_{0}^{\pi} \dfrac{\sin x \mathrm{dx}}{1+\cos ^{2} x}\end{array}\\$$

Put $$\cos x=t$$ so that $$-\sin x d x=d t$$

The limits are, when $$x=0, t=1$$ and $$x=\pi, t=-1,$$ we get

$$\begin{array}{l}I=\dfrac{-\pi}{2} \displaystyle \int_{1}^{-1} \dfrac{\mathrm{dt}}{1+t^{2}}=\pi \int_{0}^{1} \dfrac{\mathrm{dt}}{1+t^{2}}\\ \left[\because \int_{a}^{-a} f(x) \mathrm{d} \mathrm{x}=-\int_{-a}^{a} f(x) \mathrm{d} \mathrm{x} \text { and } = -2\int_{0}^{2 a} f(x) \mathrm{d} \mathrm{x} ,\ \ \text{if f(x) is even function}\right]. \\I=\pi | \tan ^{-1} t |_0^1=\pi \left[ \tan ^{-1} 1-\tan ^{-1} 0\right]=\pi\left[\dfrac{\pi}{4}-0\right]=\dfrac{\pi^{2}}{4}\end{array}\\$$
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Subjective Medium Published on 17th 09, 2020
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