How can I evaluate $\int \sin^{7/2} x \,dx$?
Actually, I know how to compute integral $\int \sin^{n} x dx $ where $n \in \mathbb N$ and I specifically chose $7/2$ I can generalize this as any $q \in \mathbb Q -\mathbb N$. Also, there is no importance of $\sin x$ I can change it with $\cos x, \tan x$
I checked the calculators and they can not compute. I have an idea to write maclaurin series of sin and take its power then integrate but I am not sure
In general integrals $\int \sin^a x \,dx$ and $\int \cos^a x \,dx$, are not expressible in terms of elementary functions: Substituting $u = \sin x$ transforms the integral to $$2 \int \frac{u^a\, du}{\sqrt{1 - u^2}} = \frac{1}{a + 1} u^{a + 1} {}_2F_1\left(\frac{1}{2}, \frac{a + 1}{2}; \frac{a + 3}{2}; u^2\right) ,$$ where ${}_2 F_1$ is the ordinary hypergeometric function.
If $a$ is a half-integer, say, $a = k + \frac{1}{2}$, $k \in \Bbb Z$, substituting $u = v^2$ transforms the integral to $$2 \int \frac{v^{2 k + 2} \,dv}{\sqrt{1 - v^4}},$$ which can be evaluated in terms of incomplete elliptic functions. For example, for $k = 3$ ($a = \frac{7}{2}$), $$\int \sin^\frac72 x \,dx = -\frac{5 \sqrt{2}}{21} F\left(\sqrt{1 + \sin x}, \frac{1}{\sqrt 2}\right) + \frac{1}{21} (6 \cos^3 x - 16 \cos x) \sqrt{\sin x} + C,$$ where $F$ is the incomplete elliptic integral of the first kind.
The integrals $\int \tan^a x \,dx$ behave differently: The substitution $w = \tan x$ transforms the integral to $$\int \frac{w^a \,dw}{1 + w^2}.$$ If $a$ is rational, say, $a = \frac{p}{q}$, the substitution $w = y^q$ rationalizes the integral, and so the integral has an elementary antiderivative: $$q \int \frac{y^{p + q - 1} \,dy}{1 + y^{2 q}} .$$ (Already for $a = \frac{1}{2}$ computing an explicit antiderivative is somewhat involved.)
For general exponents $a$, $$\int \frac{w^a \,dw}{1 + w^2} = \frac{1}{a + 1} {}_2 F_1 \left(1, \frac{a + 1}{2}; \frac{a + 3}{2}; -w^2\right) = \frac{1}{2} w^{a + 1} \Phi\left(-w^2, 1, \frac{a + 1}{2} \right) ,$$ where $\Phi$ is the Lerch transcendent function.
Incomplete beta function
$\def\B{\operatorname B}\def\F{\operatorname F}$ Since the incomplete beta function is:
$$\int\sin(x)^{2a-1}\cos(x)^{2b-1}dx= \frac12\B_{\sin^2(x)}(a,b)+C$$
we get $a=\frac94,b=\frac12$:
$$\int\sin^\frac72(x)dx=\frac12 \B_{\sin^2(x)}\left(\frac94,\frac12\right)+C$$
Using a recurrence identity and the Pochhammer symbol $(u)_v$,
$$\B_x(a+n,b)=\frac{(a)_n}{(a+b)_n}\B_x(a,b)-\frac1{a+b+n-1}\sum_{n=1}^{n-1}\frac{(1-a-n)_k(1-x)^bx^{a+n-k-1}}{(2-a-b-n)_k}$$
Setting $n=2$:
$$\int\sin^\frac72(x)dx=\frac 5{42}\B_{\sin^2(x)}\left(\frac14,\frac12\right)-\frac2{21}\sqrt{\sin(x)}\cos(x)(3\sin^2(x)+5)$$
Convert $\B_{\sin^2(x)}\left(\frac14,\frac12\right)$ into a form with the elliptic integral of the first kind with $m=k^2\ \F(x,m)$:
$$\int \sin^\frac72(x)dx=\frac1{42}\sqrt{\sin(x)}(3\cos(3x)-23\cos(x))-\frac{10}{21}\F\left(\frac\pi4-\frac x2,2\right)+C\tag1$$
Reduction formula
Another way is via reduction formula:
$$\begin{align}\int\sin^\frac72(x)=-\frac27\sin^\frac52(x)\cos(x)+\frac57\int\sin^\frac32(x)dx=\frac1{21}(3\cos(3x)-13)\sqrt{\sin(x)}\cos(x)+\frac5{21}\int\sqrt{\csc(x)}dx\end{align}$$
Finally, using $\B_x(a,b)$ on $\int\sin^{-\frac12}(x)$ and converting to $\F(x,m)$, or a similar method, again gives $(1)$.
