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Question Number 118340 by bramlexs22 last updated on 17/Oct/20

\underset{π / 3}{\int^{π / 2}} \frac{d x}{1 + \sin x - \cos x}

Commented by Dwaipayan Shikari last updated on 17/Oct/20

\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d x}{1 - c o s x + s i n x} = 2 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{1 - \frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} . \frac{1}{1 + t^{2}} x = t a n \frac{x}{2}

= 2 \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{2 t^{2} + 2 t} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t (t + 1)} = {[l o g (\frac{t}{t + 1})]}_{\frac{1}{\sqrt{3}}}^{1}

= l o g (\frac{1}{2}) - l o g (\frac{1}{\sqrt{3} + 1}) = l o g (\frac{\sqrt{3} + 1}{2})

Answered by Bird last updated on 17/Oct/20

I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d x}{1 + s i n x - c o s x}

w e d o t h e c h . t a n (\frac{x}{2}) = t \Rightarrow

I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 d t}{(1 + t^{2}) (1 + \frac{2 t}{1 + t^{2}} - \frac{1 - t^{2}}{1 + t^{2}})}

= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 d t}{1 + t^{2} + 2 t - 1 + t^{2}}

= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 d t}{2 t^{2} + 2 t} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t (t + 1)}

= \int_{\frac{1}{\sqrt{3}}}^{1} (\frac{1}{t} - \frac{1}{t + 1}) d t

= {[l n ∣ \frac{t}{t + 1} ∣]}_{\frac{1}{\sqrt{3}}}^{1} = l n (\frac{1}{2}) - l n (\frac{1}{\sqrt{3} (1 + \frac{1}{\sqrt{3}})})

= - l n (2) + l n (1 + \sqrt{3})

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