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Question

Evaluate :

$$\displaystyle\int_{0}^{\pi} \dfrac{1}{1+\sin x} dx$$


Solution

$$\displaystyle \int_{0}^{\pi } \frac{1}{1+sinx}dx $$
$$\displaystyle = \int_{0}^{\pi } \frac{1-sinx}{1-sin^{2}x}dx $$ Raticnaising the devotion 
$$\displaystyle = \int_{0}^{\pi } \frac{1-sinx}{cos^{2}x}dx $$
$$\displaystyle = \int_{0}^{\pi } (sec^{2}x-tan\,x sec\,x) dx $$
$$\displaystyle | tan x - sec\,x |_{0}^{\pi } $$
$$\displaystyle = tan \pi -tan\,0 - sec\pi +sec\,0 $$
$$\displaystyle = 0-0-(-1)+1 $$
$$ \Rightarrow 1+1 = 2 $$ 


1120315_1140112_ans_fb3cb06a1d7b42a18edbdd37c91b4bac.jpg

Mathematics

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