A pair of rational power integrals $ \int (\sin x + 1)^{1/2}dx, \ \int (\sin x + 1)^{3/2}dx$

Question

I'm currently teaching an introductory calculus course which goes through various "techniques of integration." On the way to showing that we can integrate $$ \int R(x, \sqrt{ax^2 + bx + c})dx $$ for a rational function $R(u,v)$, we teach trig substitution, partial fraction decomposition, and then we deal with integrals of the form $$ \int \sin^n(x)\cos^m(x) dx \quad \text{or} \quad \int\tan^n(x) \sec^m(x) dx $$ for integral powers of $n$ and $m$. While looking at another problem here, I began to wonder about non-integer powers of sine and cosine. A few quick scribbles later, I found that these are very hard very quickly. Often elliptic integrals and hypergeometric functions came up, which I took as indicators that I should look at something different.

But then I happened to cross upon the integrals $$ \int (\sin x + 1)^{1/2}dx \quad \text{and} \quad \int (\sin x + 1)^{3/2}dx, $$ which both have nice antiderivatives that are elementary, except they include $(\sin x + 1)^{1/2}$. The antiderivative of the former is $$ \frac{2\sqrt{\sin x + 1}\left( \sin\frac{x}{2}- \cos\frac{x}{2} \right)}{\sin\frac{x}{2} + \cos\frac{x}{2} } + C. $$ The antiderivative of the second is of a similar flavor, but more annoying to write down. Further, other powers of the form $n/2$ have similar flavor, but are increasingly complicated.

I'm interested in knowing a good method of evaluating these integrals, whether or not there is any reference or source that deals with this type of integral, and whether or not there is any reference that deals with other generalizations of integrals of the form $\int \sin^n x \cos ^m x dx$.

Answer 1

Below is a systematic approach. Note that $\sin x+1=(\sin\frac x2+\cos\frac x2 )^2$ and $$ \int (\sin x + 1)^{\frac{2m+1}2}dx =\text{sgn} (\sin x + 1)J_m$$ where $J_m$ admits a reduction formula \begin{align} J_m=& \int (\sin\frac x2+\cos\frac x2)^{2m+1}dx\\ =&\ \frac{(\sin x+1)^{m}}{m+\frac12}(\sin\frac x2-\cos\frac x2)+\frac{2m}{m+\frac12}J_{m-1} \end{align} with $ J_0=2(\sin\frac x2-\cos\frac x2)$. Express $\text{sgn} (\sin x + 1)=\frac{\sqrt{\sin x+1}}{\sin\frac x2+\cos\frac x2}$ to successively obtain \begin{align} \int (\sin x + 1)^{\frac12}dx =& \ \text{sgn} (\sin x + 1)J_0\\ = &\frac{2\sqrt{\sin x+1}\ (\sin\frac x2-\cos\frac x2)}{\sin\frac x2+\cos\frac x2} \\ \int (\sin x + 1)^{\frac32}dx =&\ \text{sgn} (\sin x + 1)J_1\\ =& \frac{2\sqrt{\sin x+1}\ (\sin\frac x2-\cos\frac x2)}{3(\sin\frac x2+\cos\frac x2)}\left(\sin x+5\right)\\ \int (\sin x + 1)^{\frac52}dx =&\ \text{sgn} (\sin x + 1)J_2= \cdots \\ \end{align}

Answer 2

$\sin x=\cos\bigg(\dfrac\pi2-x\bigg)$, and $1+\cos t=2\cos^2\dfrac t2$. Then we are left with an integral of the form $\displaystyle\int\cos^ku~du$. Now, if $k=2p+1$ is odd, this becomes $\displaystyle\int\big(1-\sin^2u\big)^p~d(\sin u)$, which is trivial. Else, if $k=2p$ is even, establish a recurrence relation between $I\big(2p\big)$ and $I\big(2p-2\big)$. Similarly for $\displaystyle\int\sin^nx~cos^mx~dx$, where, if either m or n is odd, then a simple substitution renders the integral trivial. Else, when both m and n are even, a recurrence relation is established. Hope this helps.