Mathematics

Evaluate:
$$\displaystyle \int_{0}^{\pi} \dfrac{x}{1 + sinx}$$


SOLUTION
Let $$I = \displaystyle \int_{0}^{\pi} \dfrac{x}{1 + sin x}dx$$ ....(i)
and $$I = \displaystyle \int_{0}^{\pi} \dfrac{\pi - x}{1 + sin(\pi - x)}dx = \displaystyle \int_{0}^{\pi} \dfrac{\pi - x}{1 + sin x}dx$$ .....(ii)

On adding Eqs. (i) and (ii), we get
$$2I = \pi \displaystyle \int_{0}^{\pi} \dfrac{1}{1 + sin x}dx$$
$$= \pi \displaystyle \int_{0}^{\pi} \dfrac{(1 - sin \,x)dx}{(1 + sin \,x)(1 - sin\,x)}$$
$$= \pi \displaystyle \int_{0}^{\pi} \dfrac{(1 - sin \,x)dx}{cos^2x}$$
$$= \pi \displaystyle \int_{0}^{\pi} (sec^2 x - tan \,x . sec\,x)dx$$
$$= \pi \displaystyle \int_{0}^{\pi} sec^2 xdx - \pi \displaystyle \int_{0}^{\pi} sec \,x \,tan \,x \,dx$$
$$= \pi[ (tan\,x)_{0}^{\pi} - (sec \,x)_{0}^{\pi}]$$
$$= \pi (tan \,pi - tan \,0 - sec \pi + sec \,0)$$
$$\Rightarrow 2I = \pi (0 - 0 + 1 + 1) = 2 \,\pi$$
$$2I = 2 \pi$$
$$\therefore I = \pi$$
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Subjective Medium Published on 17th 09, 2020
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