Integrating cos and csc

Question

$$\int{\frac{\sqrt{\cos x+1}}{\csc x}dx}$$ Sorry for the bad formatting, I still need to learn math jax. I am trying to integrate this by u-sub but am stuck. Looking at the equation I can see somewhere I will probably have to use the fact the $\csc x=1/\sin x$. My hunch is I choose $u$ to be something in the numerator, but is there an identity i'm not seeing?

Answer 1

Guide:

\begin{align} \int \frac{(\cos x+1)^\frac12}{ \csc x} \, dx = \int \sin x(\cos x+1)^\frac12 \, dx \end{align}

Try substitution $u = \cos x + 1$.

Answer 2

Here's my solution

$$\int{\sqrt{(cos(x)+1)}\over(csc(x))}dx$$ Which gives $$\int{\sqrt{(cos(x)+1)}*sin(x)}dx$$ Then let $u=cos(x)+1$ which means $-du=sin(x)$ $$-\int{\sqrt{u}}du$$ $$-{2u^{3/2}\over{3}}+C$$ Which gives $$-{2(cos(x)+1)^{3/2}\over{3}}+C$$

Answer 3

You may use double angle formulas first to simplify:

• $1+\cos{x} = 2\cos^2 \frac{x}{2}$
• $\frac{1}{\csc x}= \sin x = 2\cos\frac{x}{2}\sin\frac{x}{2}$

$$\int{\frac{\sqrt{\cos x+1}}{\csc x}dx}=2\sqrt{2}\int \cos^2\frac{x}{2} \cdot\sin\frac{x}{2}\, dx = -4\sqrt{2}\int\cos^2\frac{x}{2}\,d\left(\cos\frac{x}{2}\right)= $$$$ =-\frac{4}{3}\sqrt{2}\cos^3\frac{x}{2} + C$$