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$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$

$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$

I tried and got the answer but my answer is not matching the options given.Is my method not correct?

$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=\int\limits_{-1/2}^{1/2}(3\sin^{-1}(x)-3\cos^{-1}(x))dx$
$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$

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  • $\begingroup$ $\arcsin(a+b)\neq\arcsin a+\arcsin b$. $\endgroup$
    – Intelligenti pauca
    Aug 9, 2015 at 14:08
  • $\begingroup$ @Aretino,where have i used that? $\endgroup$
    – diya
    Aug 9, 2015 at 14:10
  • $\begingroup$ The substitution $3x-4x^{3} \to x$ is incorrect $\endgroup$
    – Leucippus
    Aug 9, 2015 at 14:15
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    $\begingroup$ @diya Sorry, I had misread your work. However your change of variables in the integral is incorrect. $\endgroup$
    – Intelligenti pauca
    Aug 9, 2015 at 14:41
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    $\begingroup$ I think $\arccos(4x^3-3x)$ is not $3\arccos(x)$ in this interval. For the angle $\arccos(x)$ is not far from $90$ degrees, so multiplying by $3$ we get something much bigger than $\arccos(4x^3-3x)$. $\endgroup$
    – André Nicolas
    Aug 9, 2015 at 14:42

2 Answers 2

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Hint:

$$\arcsin{a} - \arccos{(-a)} = \arcsin{a}+\arcsin{(-a)} - \frac{\pi}{2} = - \frac{\pi}{2}$$

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We use the facts that $$\arccos(-x)=\pi-\arccos(x)$$ and $$\arccos(x)+\arcsin(x)=\pi/2.$$ We have \begin{align} & \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\arccos\left(4x^3-3x\right)\right)\mathrm{d}x \\ & \quad = \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\pi+\arccos\left(3x-4x^3\right)\right)\mathrm{d}x \\ & \quad = \int_{-1/2}^{1/2}\left(-\pi+\pi/2\right)\mathrm{d}x=-\pi/2. \end{align}

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  • $\begingroup$ @Leucippus Thanks for the edit, it looks better now! I was hurry because of Ron Gordon's answer (+1 btw for another trig identity). $\endgroup$
    – Nicolas
    Aug 9, 2015 at 14:19
  • $\begingroup$ just trying to help, also good, clear and instructive answer. $\endgroup$
    – Leucippus
    Aug 9, 2015 at 14:21
  • $\begingroup$ Why on earth would my answer cause you to hurry yours? $\endgroup$
    – Ron Gordon
    Aug 9, 2015 at 14:29
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    $\begingroup$ @Ron Gordon I meant that I saw your answer while finishing writing mine, so I did not pay attention to the "elegance" of my answer. I do not answer when I see someone has posted something similar to what I want to write. $\endgroup$
    – Nicolas
    Aug 9, 2015 at 14:39

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