How do you get the integrals for $\sqrt{1 + \sin{x}}$ and $\sqrt{1 + \cos{x}}$?

Question

integral-calculator.com gets $2 \sin{(\frac{x}2)} - 2 \cos{(\frac{x}2)} + C$ for the indefinite integral of $\sqrt{1 + \sin{x}}$, and $2^{\frac{3}2}\sin(\frac{x}2)+C$ for that of $\sqrt{1 + \cos{x}}$. However, the method it uses involves these advanced trig identities which I've never heard of:

...and my professor has literally warned me about unorthodox derivations with online integral calculators.

Answer 1

This is due to half-angle identities, for example $1+\cos x=2\cos^2\left(\frac{x}{2}\right)$. This half-angle formula is a special case rewriting and rearrangement of the double angle formula $\cos(2\theta)=2\cos^2(\theta)-1$, by plugging in $\theta=\frac{x}{2}$, and adding $1$ to both sides of the equation. And the double-angle formula is itself a special case of the addition formula $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ (with $\alpha=\beta=\theta$), and the pythagorean identity $\sin^2+\cos^2=1$.

Summary: addition formula + pythagorean identity $\implies$ double-angle formula $\implies$ half-angle formula.

Answer 2

As Doug said in his comment, the identity $\cos(2x)+1 = 2\cos^2x$ is being used here.

$$\sqrt{\cos x +1} = \sqrt{2\cos^2(\frac{x}{2})} = \sqrt{2}\cos(\frac{x}{2})$$

From here on, a $u = \frac{x}{2}$ substitution should suffice to solve the integral itself.

Answer 3

Hint:

$$\int \sqrt{1+\sin x}~ dx=\int \sqrt{\sin(x/2)+\cos(x/2)}~ dx=\int |\sin(x/2)+\cos(x/2)]dx+C_1$$ $$\int \sqrt{1+\cos x}~ dx=\int \sqrt{2\cos ^2(x/2)}~ dx= \int \sqrt{2}~ |\cos(x/2)|~ dx+C_2.$$

Actually, the upper and lower limits of the definite version of these integrals helps removing the signs $|.|$ to get the final answer.