2
$\begingroup$

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

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:

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

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

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

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

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

$\endgroup$
7
  • 2
    $\begingroup$ That's not the derivative of arctan. $\endgroup$ – Unit May 9 '19 at 23:40
  • 5
    $\begingroup$ The second (awful) part of the sum is an odd function, so its integral on a symetric interval is $0$. $\endgroup$ – Bernard Masse May 9 '19 at 23:46
  • 1
    $\begingroup$ @BernardMassé: You should write that as an answer. Your comment essentially solves the problem. $\endgroup$ – Clayton May 9 '19 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. $\endgroup$ – Bernard Masse May 10 '19 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 '19 at 0:41
3
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ woooooooah that was nice (+1) $\endgroup$ – clathratus May 10 '19 at 0:45
  • $\begingroup$ @clathratus thanks! $\endgroup$ – coreyman317 May 10 '19 at 1:21
  • 1
    $\begingroup$ Yesssss! thank you so much :) $\endgroup$ – Brandy May 10 '19 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.