TITLE: $\int_0^{1/2} \cos^{-1} x \, dx$ using integration by parts.
QUESTION [2 upvotes]: integral lower bound is 0. upper bound is 1/2. function is cos^-1 x dx.
My work: I use integration by parts. 
u = cos^-1 x ...
du = -dx/sqrt(1-x^2) ... 
v = x ...
dv = dx ...
so integral udv = [1/2cos^-1(1/2) - 0(cos^-1) 0 + \int 0 ^ 1/2 xdx/(sqrt(1-x^2) = 1/2* pi/3 + (1/2)dx/ sqrt(1-x^2) - 0dx/sqrt(1-x^2).
Is the last part after pi/3 on the right track? For some reason, the textbook had lower bound 1 and upper bound 3/4 on the intergal t^-1/2 [-1/2dt] instead?

REPLY [2 votes]: Using integration by parts as you did, setting $u = \cos^{-1}(x)$ and $dv = dx \implies v = x$, your indefinite integral should be
$$uv - \int v\,du = x\cos^{-1}x - \int \underbrace{x}_{v}\underbrace{\left(\frac{-\,dx}{\sqrt{1-x^2}}\right)}_{du}$$
It looks like you managed the integration by parts okay (at least the set up, even though you refer to $\int u\,dv \;?$. However, it seems you never integrated the integral on the right before evaluating.
To integrate, you can do this easily by substitution $t = 1 - x^2 \iff dt = -2x \iff \frac 12 dt = -x\,dx)$. 
The bounds on this right-side integral change under this substitution: when $\;x = 1/2,\; t = 1-(1/2)^2 = \frac 34.\;$ When $\;x = 0$, $\;t = 1$.
$$\begin{align}\int_0^{1/2} \cos^{-1} x \, dx & = x \cos^{-1}x\Big|_0^{1/2} - \frac 12 \int_{1}^{3/4}t^{-1/2}\,dt\,\\ \\ & = x \cos^{-1}x\Big|_0^{1/2} - t^{1/2}\Big|_{1}^{3/4} \end{align}$$
{"set_name": "stack_exchange", "score": 2, "question_id": 709456}