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$$\int_{-\pi/4}^{\pi/4}\bigl(\cos x+ \sqrt {1+x^2}\sin^3x \cos^3x \bigr)\, dx $$

This question is from a math GRE practice test

I've tried to solve this integral for 2 days... starting to think it is a typo

The answer is $\sqrt2$ but I need to know how to solve with steps

Some useful trig identities and derivatives:

$$\frac{d}{dx} \sin^2x = 2\cos x \sin x$$

$$\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$$

$$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x$$

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

$$\sin 2x = 2\cos x \sin x$$

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    $\begingroup$ That's not the derivative of arctan. $\endgroup$
    – Unit
    May 9, 2019 at 23:40
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    $\begingroup$ The second (awful) part of the sum is an odd function, so its integral on a symetric interval is $0$. $\endgroup$
    – MasB
    May 9, 2019 at 23:46
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    $\begingroup$ @BernardMassé: You should write that as an answer. Your comment essentially solves the problem. $\endgroup$
    – Clayton
    May 9, 2019 at 23:49
  • $\begingroup$ @Clayton: Why should I give an answer? The hint is sufficient for the person who asked for help to finish the work. This is how I tried to help students in my teaching career: either answer a question with another (helpful) question or give just enough (and not more) information to finish the problem. In this case, I think I shouldn't even have written the value of the integral. $\endgroup$
    – MasB
    May 10, 2019 at 0:00
  • $\begingroup$ @BernardMassé: I didn't say that you should give a solution. I said that you should write your comment (verbatim) as an answer so that it can accumulate upvotes. I think you read too much into my comment. The hint is plenty sufficient and I think is was well said; I gave the comment an upvote and would have provided one for an answer as well. $\endgroup$
    – Clayton
    May 10, 2019 at 0:41

1 Answer 1

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Let $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\cos(x)+\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$

For the second integral, notice $$f(x)=\sqrt{1+x^2}\sin^3(x)\cos^3(x)$$ is an odd function since $$f(-x)=\sqrt{1+(-x)^2}\sin^3(-x)\cos^3(-x)=-\left(f(x)\right)$$

So recall that $$\int_{-b}^bf(x)dx=0$$ if $f$ is odd and the integral exists over the interval. Conclude that $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+x^2}\sin^3(x)\cos^3(x)dx=0.$

Then $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+0=2\int_0^{\frac{\pi}{4}}\cos(x)dx=2\sin(x)\bigg|^{\frac{\pi}{4}}_0=2\sin\left(\frac{\pi}{4}\right)-2\sin(0)=\sqrt2$$

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  • $\begingroup$ woooooooah that was nice (+1) $\endgroup$
    – clathratus
    May 10, 2019 at 0:45
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    $\begingroup$ Yesssss! thank you so much :) $\endgroup$
    – Brandy
    May 10, 2019 at 3:33

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