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So I was doing a integral question and I stumbled upon this question.

$\displaystyle{\int\sqrt{1 + \sin\left(\frac x2\right)}\,dx}$

In order to solve it I did the following:

I took $u = \frac12x$ Then $\frac {du}{dx}$ Which gave me $2 du = dx$.

After that I substituted u in the equation to get $2\int \sqrt{1 + \sin(u)} du$.

After this I was stuck as I am new to integration of trigonometry so I checked my textbook which did the same just the same but the step after this was this one

$2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$

I am in a complete awe how the textbook got $2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$ from $2\int \sqrt{1 + \sin(u)} \text{ du} $

Can someone please explain me how the this is achieved? I am totally stuck

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    $\begingroup$ Pythagorean theorem and double angle formula for the sine $\endgroup$ – imranfat Aug 12 at 16:52
  • $\begingroup$ You have $1 = \sin^2 \alpha + \cos^2 \alpha$ and also $\sin 2 \alpha = 2 \sin \alpha \cos \alpha$. Sure, I'd be staring at the question for a solid month without seeing that one. $\endgroup$ – vonbrand Aug 12 at 17:04
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First, they used the double angle formula:

$$\sin u = \sin\left(2\frac{u}{2}\right) = 2\sin\frac{u}{2}\cos\frac{u}{2}.$$

Then they replaced the $1$ with $\sin^2\frac{u}{2}+\cos^2\frac{u}{2}.$

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  • $\begingroup$ Oh Ya I totally forgot. Thank You for your help :) $\endgroup$ – Utkarsh Aug 12 at 16:58
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Use $$\sqrt{1+\sin\frac{x}{2}}=\sqrt{1+\cos\left(\frac{\pi}{2}-\frac{x}{2}\right)}=\sqrt2\left|\cos\left(\frac{\pi}{4}-\frac{x}{4}\right)\right|$$

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    $\begingroup$ @mrtaurho Thank you for your editing. $\endgroup$ – Michael Rozenberg Aug 12 at 16:57
  • $\begingroup$ is it same as the answer $4[\sin\frac14x - \cos\frac14x] + c$ as this is the answer in my textbook.. $\endgroup$ – Utkarsh Aug 12 at 16:57
  • $\begingroup$ @Utkarsh Use also $\sqrt2\left(\cos\left(\frac{\pi}{4}-\frac{x}{4}\right)\right)=\cos\frac{x}{4}+\sin\frac{x}{4}$ $\endgroup$ – Michael Rozenberg Aug 12 at 17:00
  • $\begingroup$ Oh Okay got it. Thank You :) $\endgroup$ – Utkarsh Aug 12 at 17:01
  • $\begingroup$ @Utkarsh You are welcome! $\endgroup$ – Michael Rozenberg Aug 12 at 17:01
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Obviously, knowing the trigonometric identities as already covered is the far superior and simpler method. Here is a more 'generalised' approach. I hope it serves some value.

Here I will address the integral: \begin{equation} I = \int \sqrt{1 + \sin\left(\frac{x}{2}\right)}\:dx \end{equation}

First let $u = \frac{x}{2}$:

\begin{equation} I = \int \sqrt{1 + \sin(u)} \cdot 2\:du = 2\int \sqrt{1 + \sin(u)}\:du \end{equation}

We now employ the Weierstrauss Substitution $t = \tan\left(\frac{u}{2}\right)$: \begin{align} I &= 2 \int \sqrt{1 + \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt = 4 \int \frac{t + 1 }{\left(t^2 + 1 \right)^{\frac{3}{2}}} \:dt = 4\left[ \int \frac{t}{\left(t^2 + 1\right)^{\frac{3}{2}}}\:dt+ \int \frac{1}{\left(t^2 + 1\right)^{\frac{3}{2}}}\:dt \right] \nonumber ]\\ &= 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + J\right] \end{align}

For the remaining integral $J$ let $t = \tan(s)$: \begin{align} J &= \int \frac{1}{\left(\tan^2(s) + 1\right)^{\frac{3}{2}}} \cdot \sec^2(s)\:ds = \int \cos(s) \:ds = sin(s) + C = \sin\left(\arctan(t)\right) + C \end{align} Where $C$ is the constant of integration.

Thus, \begin{equation} I = 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + J\right] = 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + \sin\left(\arctan(t)\right)\right] + C \end{equation}

Now $t = \tan\left(\frac{u}{2} \right) = \tan\left(\frac{x}{4}\right)$

Thus, \begin{align} I &= 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + \sin\left(\arctan(t)\right) \right] + C = 4\left[ -\frac{1}{\sqrt{\tan^2\left(\frac{x}{4}\right) + 1}} + \sin\left(\arctan\left(\tan\left(\frac{x}{4}\right)\right)\right) \right] + C \\ &= 4\left[ -\cos\left(\frac{x}{4} \right) + \sin\left(\frac{x}{4} \right) \right] = 4\cdot \sqrt{2}\sin\left(\frac{x}{4} = \frac{\pi}{4} \right) + C \end{align}

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