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Problem Statement: What is the value of $\int_{-\pi/4}^{\pi/4}(\cos t +\sqrt{1+t^{2}}\sin^{3}t\cos^{3}t)dt?$

I am working on an old GRE Math Subject test, and I am having trouble determining which substitution to do for the given integral.

With algebraic substitution, I have $$u = \cos t,\ \ du = -\sin t dt$$ with the resulting integral $$\sin t - \int \sqrt{1+\arccos^{2}u}\ u^{3}(1-u^{2})du.$$ This seems not to simplify things because of the $\sqrt{1+\arccos^{2}u}$ factor.

With trigonometric substitution, we let $$t = \tan\theta,\ \ dt=\sec^{2}\theta d\theta$$ with the resulting integeral $$\sin t + \int \sec^{3}\theta\ \sin^{3}(\tan\theta)\cos^{3}(\tan\theta)d\theta.$$ Then if I try to do an algebraic-substitution from here I believe I will result in the original integral...

Should I be using integration-by-parts for this integral? I did not try this method because I figured it would be far too complicated, if even possible...

Any suggestions for how I should tackle this integral? The correct resulting value is $\sqrt{2}$.

Thank you

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1 Answer 1

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The second part is an odd function and integrates to 0 on $[-\pi/4, \pi/4]$.

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  • $\begingroup$ aha! I thought that maybe odd/even had something to do with it! So since $\sin t$ is odd, $\sin^{3} t$ is odd, and since $\cos t$ is even, $\cos^{3} t$ is even, and since $\sqrt{1+t^{2}}$ is even, $\sqrt{1+t^{2}}\cos^{3} t$ is even. Thus, $\sqrt{1+t^{2}}\sin^{3} t\cos^{3} t$ is odd... correct? $\endgroup$
    – yung_Pabs
    Mar 17, 2016 at 1:52
  • $\begingroup$ That is a correct argument. $\endgroup$ Mar 17, 2016 at 1:53
  • $\begingroup$ Awesome! I will need to memorize the properties of even/odd functions... I had a feeling actually computing that integral would be way too complicated... thank you! $\endgroup$
    – yung_Pabs
    Mar 17, 2016 at 1:55
  • $\begingroup$ @yung_Pabs I feel sorry that you have to do the silly GRE. Wish you all the luck! $\endgroup$
    – imranfat
    Mar 17, 2016 at 2:55
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    $\begingroup$ @yung_Pabs Yes I am aware of that (silly) requirement. You must be in the USA I presume... $\endgroup$
    – imranfat
    Mar 18, 2016 at 21:31

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