Applying substitution to $\int \sqrt{1+\sin(x)}$

Question

I'm having a problem with substituting when it comes to trig functions.

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

Substituting: $1+\sin(x) = u \implies \frac{d}{dx} u = \cos(x) \implies dx = \frac{1}{\cos(x)}du$

So now we have the integral: $$\int \frac{\sqrt{u}}{\cos(x)}du$$

So now the question is, what do I do with the $\frac{1}{\cos(x)}$? How can I substitute that with $u$?

Answer 1

Hint Write your integrand in the form $$\sqrt{\frac{(1+\sin(x))(1-\sin(x))}{1- \sin(x)}}=\frac{\pm\cos(x)}{\sqrt{1-\sin(x)}}$$ and substitute $$t=1-\sin(x)$$ then we get $$dt=-\cos(x)dx$$

Answer 2

HINT

Here I give you an alternative approach. Observer that

\begin{align*} 1 + \sin(x) & = 1 + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = \left[\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)\right]^{2} \end{align*}

Hence the given integral is equal to

\begin{align*} \int\sqrt{1+\sin(x)}\mathrm{d}x = \int\left|\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)\right|\mathrm{d}x \end{align*}

Answer 3

Hint: $$ \begin{align} \sqrt{\frac{1+\sin(x)}2} &=\sqrt{\frac{1+\cos\left(\frac\pi2-x\right)}2}\\ &=\left|\,\cos\left(\tfrac\pi4-\tfrac x2\right)\,\right| \end{align} $$