veceriraby

2022-01-23

Evaluating ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left(\mathrm{sin}x+2\mathrm{cos}x\right)}^{3}dx$

Damian Roberts

Expert

Just expand the binomial
${\left(\mathrm{sin}x+2\mathrm{cos}x\right)}^{3}={\mathrm{sin}}^{3}x+6{\mathrm{sin}}^{2}x\mathrm{cos}x+12\mathrm{sin}x{\mathrm{cos}}^{2}x+8{\mathrm{cos}}^{3}x$
Now note that
${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\mathrm{sin}}^{3}xdx=0={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}12\mathrm{sin}x{\mathrm{cos}}^{2}xdx$
(because the functions inside the integrals are odd functions and since we are in a symmetric interval). The rest of the integrals are easy, since
${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(6{\mathrm{sin}}^{2}x\mathrm{cos}x+8{\mathrm{cos}}^{3}x\right)dx={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(-2{\mathrm{sin}}^{2}x+8\right)\mathrm{cos}xdx$
and if we set $u=\mathrm{sin}x$, the integral is much easier :
${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(-2{\mathrm{sin}}^{2}x+8\right)\mathrm{cos}xdx={\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left(-2{u}^{2}+8\right)du=\frac{23\sqrt{2}}{3}$

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