Anti-derivative of $\frac{dy}{dx}=(1+sin(x))^a$ where $a$ is a positive real number

Question

Find the anti-derivative of this function where $a>0$:$$\frac{dy}{dx} =(1+sin(x))^a$$ This is the answer given by Wolfram (link):

$$ y = \frac{\sqrt{2}\, cos(x) (sin(x) + 1)^a\, _2F_1(\frac{1}{2}, a + \frac{1}{2}, a + \frac{3}{2}, \frac{cos^2(x)}{2 - 2 sin(x)})}{(2 a + 1) \sqrt{1 - sin(x)}} + constant$$

But this has singularities whenever $sin(x)=1$, irrespective of the value of $a$, which says to me that it is not correct.

By comparison, the anti-derivative of $y=(1+sin(x))^1$ given by Wolfram is entirely different to the above result at $a=1$.

Answer 1

The result $$y=\int (1+\sin(x))^a\,dx=\frac{\sqrt{2} \cos (x) (1+\sin (x))^a }{(2 a+1) \sqrt{1-\sin (x)}}\, _2F_1\left(\frac{1}{2},a+\frac{1}{2};a+\frac{3}{2};\frac{\cos ^2(x)}{2-2 \sin (x)}\right)+C$$ seems to be correct (according to other CAS).

Computing $$y=\int_0^t (1+\sin(x))^a\,dx$$ this simplifies to $$y=-2^a \left(B_{\frac{1}{2}}\left(a+\frac{1}{2},\frac{1}{2}\right)-\sqrt{\cos ^2(t)} \sec (t) B_{\frac{1}{2} (1+\sin (t))}\left(a+\frac{1}{2},\frac{1}{2}\right)\right)$$ where appears the incomplete beta function.

Comparing to the standard calculations for small positive integer values of $a$, the results perfectly match of $0 \leq t \leq \frac \pi 2$. For sure, at $t=\frac \pi 2$, the formula shows a discontinuity.

Answer 2

Subsutituting $x=\frac{\pi}{2}-z$

we get, $dx=-dz$

Therefore,

$y = - \int{(1+\cos z)^a} $ $dz$

$ = - \int{2^a \cos^{2a} z/2} $ $dz$

Substituting $2a=b$ and $z/2=w$

we get $dz=2dw$

$y = -2^{b/2}*2 \int{\cos^bw} $ $dw$

$= -2^{a+1}*$ I

If $a$ is an integer, the integral $I$ can be calculated using the well-known reduction formula.

See: Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$

https://en.wikipedia.org/wiki/Integration_by_reduction_formulae

Otherwise, $I$ can be expressed using the hypergeometric function.

https://www.wolframalpha.com/input/?i=integrate+y%3D(cos+w)%5Eb,+b%3E0

Substituting $w=\pi/4-x/2$ to get

$y= 2^{a+1} \frac{\sqrt{\sin^2 (\pi/4-x/2)} \csc(\pi/4-x/2) \cos^{2a+1} (\pi/4-x/2) _2F_1 (1/2; (2a+1)/2;(2a+3)/2;\cos^2 (\pi/4-x/2))}{2a+1} $ + constant