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
$$I = \int \sqrt{1+\sin x}dx$$
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 $$ I = 2\left(-\frac{1}{\sqrt{1+u^2}}+\frac{u}{\sqrt{1+u^2}}\right) +C$$