Question

integrate 1/(1 - sin x) dx from - π/4 to π/4


please write answer for 3marks

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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\dfrac{dx}{1-sin(x)}}$

$\displaystyle\rm{=\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\dfrac{dx}{2\,sin^2\left(\dfrac{x}{2}\right)}}$

$\displaystyle\rm{=\dfrac{1}{2}\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\,cosec^2\left(\dfrac{x}{2}\right)\,dx}$

$\bf{Put\,\,\,\dfrac{x}{2}=t}$

$\bf{\implies\dfrac{dx}{2}=dt}$

$\displaystyle\rm{=\int^{\frac{\pi}{8}}_{-\frac{\pi}{8}}\,cosec^2(t)\,dt}$

$\displaystyle\rm{=-\Big[cot(t)\Big]^{\frac{\pi}{8}}_{-\frac{\pi}{8}}}$

$\rm{=-\left[cot\left(\dfrac{\pi}{8}\right)-cot\left(-\dfrac{\pi}{8}\right)\right]}$

$\rm{=-\left[cot\left(\dfrac{\pi}{8}\right)+cot\left(\dfrac{\pi}{8}\right)\right]}$

$\rm{=-2\,cot\left(\dfrac{\pi}{8}\right)}$