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The integral is:$$\int\sqrt{1+\sin x} dx$$

My first attempt was to multiply by: $$\frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}}$$

giving: $$\int\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin x}}dx$$ which is: $$\int\frac{\cos x}{\sqrt{1-\sin x}}dx$$ then make the substitution: $$u=1-\sin x$$ which gives $$\int\frac{-1}{\sqrt{u}}du$$ which gives: $$-2\sqrt{u}+c$$ so: $$I=-2\sqrt{1-\sin x}+c$$ but I am not sure if this is correct.

Also sorry if the formatting is a bit rubbish

Darío A. Gutiérrez
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Henry Lee
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1 Answers1

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$$I = \int \sqrt{1+\sin x}dx$$
Apply substitution $ u=\tan \left(\frac{x}{2}\right)$
(Tangent half-angle substitution) $$\int \frac{2\sqrt{\left(u+1\right)^2}}{\left(1+u^2\right)\sqrt{u^2+1}}du$$ $$2\int \frac{u+1}{\left(1+u^2\right)\sqrt{u^2+1}}du$$ $$2\left(\underbrace{\int \frac{u}{\left(1+u^2\right)\sqrt{u^2+1}}du}_{I_1}+\underbrace{\int \frac{1}{\left(1+u^2\right)\sqrt{u^2+1}}du}_{I_2}\right)$$

Set $v=1+u^2$ $$I_1 = \int \frac{1}{2v\sqrt{v}}dv = \frac{1}{2}\int v^{-\frac{3}{2}}dv =\frac{1}{2}\frac{v^{\left(-\frac{3}{2}+1\right)}}{\left(-\frac{3}{2}+1\right)} =-\frac{1}{\sqrt{v}}= -\frac{1}{\sqrt{1+u^2}}$$

By Trigonometric substitution $u=\tan \left(v\right)$ $$I_2 = \int \frac{1}{\left(u^2+1\right)^{\frac{3}{2}}}du= \int \frac{\sec ^2\left(v\right)}{\left(\tan ^2\left(v\right)+1\right)^{\frac{3}{2}}}dv = \int \frac{\sec ^2\left(v\right)}{\sec ^3\left(v\right)}dv =\int \frac{1}{\sec \left(v\right)}dv =\int \cos \left(v\right)dv =\sin \left(v\right) +C$$ $$I_2 = \sin \left(v\right) = \sin \left(\arctan \left(u\right)\right) = \frac{u}{\sqrt{1+u^2}}$$ So $$ I = 2\left(-\frac{1}{\sqrt{1+u^2}}+\frac{u}{\sqrt{1+u^2}}\right) +C$$

$$I = 2\left(-\frac{1}{\sqrt{1+\tan ^2\left(\frac{x}{2}\right)}}+\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{1+\tan ^2\left(\frac{x}{2}\right)}}\right) +C$$

Darío A. Gutiérrez
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    how do you see to make this substitution? – Henry Lee May 18 '18 at 11:34
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    @HenryLee It's pretty standard: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution The only exception is that, if the $\cos$ or $\sin$ in your integrand is always raised to an even power, use $u=\tan x$ instead. (See if you can understand why, in terms of periods.) – J.G. May 18 '18 at 11:42
  • Same comment as above: not valid on the largest interval possible (which would be $\Bbb R$). But +1 anyway as it's the standard wrong way to do it. – Jean-Claude Arbaut May 18 '18 at 12:29
  • @Jean-ClaudeArbaut Thanks for your comment and(+1). why "standard wrong way"? How would you solve it? – Darío A. Gutiérrez May 18 '18 at 12:43
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    Well, notwithstanding the fact that your derivation is not entirely satisfying ($\sqrt{(1+u)^2}=|1+u|$), the discontinuous solution that you will obtain can be adjusted by a constant on intervals where it's continuous, so as to have only removable singularities, and then you get a continuous antiderivative. Of course that may be considered overly complicated (a teacher may be asking just the Weierstrass transformation), but your result is a periodic function, while the integrand is continuous and nonnegative (thus has an antiderivative everywhere, monotonically increasing). – Jean-Claude Arbaut May 18 '18 at 12:49
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    And I write it's the "standard wrong way" beause, to be honest, I have very rarely seen this addressed correctly, and almost nobody seems to care, teachers included, computer algebra systems included. However, the link I give above shows that it can hurt badly those who don't understand the limitations of this method of integration (you can't compute a definite integral without taking care of intervals). – Jean-Claude Arbaut May 18 '18 at 12:54