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Okay my problem is:

$$ \int_0^{1/2} \arccos x \, dx $$

According to wolfram, I performed the indefinite integral correctly. Here is my math:

$$ u = \arccos x$$ $$ du = \frac{-1}{\sqrt {1-x^2}} $$ $$ v = x $$ $$dv = dx $$

---> $$ x\arccos x + \int \frac x {\sqrt{1-x^2}} \, dx $$

So at this point I move on from using the part integration technique, and I use U-sub for the remaining integral.

$$ u = -x^2 + 1 $$

$$ \frac {-1} 2 \, du = dx $$

and I end up with the final indefinite integral result of:

$$x\arccos x - \sqrt{-x^2 +1} $$

Now my problem is.. according to wolfram I am write on the money with the integral. But when I plug in my values, my answer is different from wolfram's, and the books.

From $0$ to $1/2$:

$$ \frac 1 2 \arccos \frac 1 2 - \sqrt{-(1/2)^2 + 1} = -0.604$$

$$0\arccos 0 - \sqrt{-(0)^2 + 1 } = -1 $$

$ -0.604 - (-1) = 0.395 $ but the answer is $0.657$ does anyone know why I get the bounds wrong? I'm simply plugging it in but I don't understand how my math is wrong :(

  • 1
    $arccos(1/2)/2 = \pi/6=.0.524$ and $\sqrt{3/4} = \sqrt{3}/2 = 0.866.$ Subtracting gives $-.342$, not $-.604.$ – 2017-01-08
  • 0
    Where you have $\dfrac{-1} 2 \,du=dx,$ you need $\dfrac{-1}2\,du = x\,dx. \qquad$ – 2017-01-08

1 Answers 1

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$$\int_{0}^{\frac{1}{2}}\arccos(x)dx$$ $u=\arccos(x)\implies du=\dfrac{-1}{\sqrt{1-x^2}}dx$ and $dv=dx\implies v=x$: $$uv-\int v\,du=x\arccos(x)\big\vert_0^{\frac{1}{2}}+\int_0^{\frac 1 2}\dfrac{x}{\sqrt{1-x^2}} \, dx$$ Now, $u$-sub $u=1-x^2\implies \frac{du}{-2}=x\,dx$ so that $$x\arccos(x)\big\vert_0^{\frac{1}{2}}+\int_0^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}}\,dx =x\arccos(x)\big\vert_0^{\frac{1}{2}}-\frac{1}{2}\int u^{-\frac{1}{2}}du$$ $$=x\arccos(x)\big\vert_0^{\frac{1}{2}}-\sqrt{1-x^2} \big\vert_0^{\frac{1}{2}}$$ Up to here, you've done it correctly. However, $$x\arccos(x)\big\vert_{0}^{\frac{1}{2}}=\frac{1}{2}\arccos(\frac{1}{2})-0\arccos(0)=\frac{1}{2}\cdot\frac{\pi}{3}=\frac{\pi}{6}$$ and $$\sqrt{1-x^2}\big\vert_{0}^{\frac{1}{2}}=\sqrt{\frac{3}{4}}-1$$ So, the final answer is $$\boxed{\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1}\approx 0.657573$$ Overall, your problem was simply stating that $(\frac{1}{2})\arccos(\frac{1}{2}) - \sqrt{-(\frac{1}{2})^2 + 1} = -.604$ when in fact $(\frac{1}{2}) \arccos(\frac{1}{2}) - \sqrt{-(\frac{1}{2})^2 + 1}\approx -0.342$