How can I integrate $$\sqrt{\tan x}+ \sqrt{\cot x}$$
What method should I approach with?
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1Related: http://math.stackexchange.com/questions/100253/which-is-the-easiest-way-to-evaluate-int-limits-0-pi-2-sqrt-tan-x – wythagoras Aug 14 '16 at 10:24
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Related: http://math.stackexchange.com/questions/425603/integral-int-sqrt-cot-x-dx?rq=1. The first answer evaluates this integral. – wythagoras Aug 14 '16 at 10:25
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@wythagoras thanx that helped – Vishnu JK Aug 14 '16 at 10:26
1 Answers
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Break it in $\sin x$ and $\cos x$. Then write $\sin x \cdot \cos x$ as $\sqrt{(1 - (\sin x -\cos x)^2)}$.
And then let $(\sin x - \cos x)$ as $t$.
wythagoras
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This isn't correct. First $1-(\sin x - \cos x)^2=1-(\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x)) = 2\cos(x)\sin(x)$, so $\sqrt{1-(\sin x - \cos x)^2}=\sqrt{2\cos(x)\sin(x)}$, not $\cos(x)\sin(x)$. – wythagoras Aug 14 '16 at 10:30
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@wythagoras , take sqrt 2 common outside. There must not be any issue there. – Anand Kishore Aug 14 '16 at 15:26
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It will indeed work, but the problem is that this answer is just stating plain wrong information, namely that you can rewrite $\sin(x)\cos(x)$ as $\sqrt{(1-(\sin x - \cos x)^2)}$. It should be: you can rewrite $\sqrt{\sin(x)\cos(x)}$ as $\sqrt{2(1-(\sin x - \cos x)^2)}$ (note that you aren't just missing a factor $\sqrt{2}$, you are also missing a square root. – wythagoras Aug 14 '16 at 16:06