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I stumbled on this problem and I'm stuck.

Any hints on how I could approach evaluating this particular integral?

$$ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2}\sin^3{t}\cos^3 {t})\,dt}$$

What I've tried prior to asking is using $\sin{t}\cos{t} = \frac{1}{2}\sin{2t}$ and $\sin^2{t} = 1 - \cos^2{t}$. I end up with:

$$ \int_{-\pi/4}^{\pi/4}{ (\cos{t} + \frac{1}{8}\sqrt{1 + t^2}\sin{2t} - \frac{1}{16}\sqrt{1 + t^2}\sin{4t}\cos{2t})\,dt}$$

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  • $\begingroup$ What about using the exponential representation for Sine and Cosine? $\endgroup$ – Turing Jan 31 '16 at 17:40
  • $\begingroup$ You could split the integral into two terms. The first term should be easy to integrate. For the second, you may find $\frac{1}{2}\sin(2t)=\sin(t)\cos(t)$ useful. $\endgroup$ – Pixel Jan 31 '16 at 17:40
  • $\begingroup$ Or even better: fist collect a $\cos t$ out, and then use the relationship $\cos^2 t = 1 - \sin^2 t$ to split the integral into three terms. Then you may use $\sin(t)\cos(t) = \frac{1}{2}\sin(2t)$ and again, maybe, the exp form. $\endgroup$ – Turing Jan 31 '16 at 17:42
  • $\begingroup$ The answer is: $\sqrt{2}$ $\endgroup$ – Jan Eerland Jan 31 '16 at 17:50
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Note that $$\int_{a}^{b}{f(t)}dt=\int_{a}^{b}{f(a+b-t)}dt$$

Here $$ I=\int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2}\sin^3{t}\cos^3 {t})}dt=\int_{-\pi/4}^{\pi/4}{ (\cos{(\pi/4-\pi/4-t)} + \sqrt{1 + (\pi/4-\pi/4-t)^2}\sin^3{(\pi/4-\pi/4-t)}\cos^3 {(\pi/4-\pi/4-t)})}dt$$ $$=\int_{-\pi/4}^{\pi/4}{ (\cos{t} - \sqrt{1 + t^2}\sin^3{t}\cos^3 {t})}dt$$

Thus

$$2I=\int_{-\pi/4}^{\pi/4}{ (2\cos{t})}dt$$

which can be easily solved.

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  • $\begingroup$ that was sweet. thanks! $\endgroup$ – Rayhunter Jan 31 '16 at 18:13
  • $\begingroup$ Very nifty !... $\endgroup$ – Pixel Jan 31 '16 at 19:15
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You could split the integral into two terms $$\int_{-\pi/4}^{\pi/4}{ (\cos{t} + \sqrt{1 + t^2}\sin^3{t}\cos^3 {t})\,dt}=2\int_0^{\pi\over 4}\cos{t}+0 =2\times\Big[\sin(t)\Big]_{0}^{\pi/4}=2\times\frac{1}{\sqrt{2}}=\sqrt{2}.$$

because $t \to \sqrt{1 + t^2}\sin^3{t}\cos^3 {t}~$ is an odd function.

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    $\begingroup$ This is the best method. $\endgroup$ – egreg Jan 31 '16 at 22:27

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