Question-173874

Question Number 173874 by akolade last updated on 20/Jul/22

Answered by a.lgnaoui last updated on 21/Jul/22

I = \int_{π / 4}^{15 π / 4} (\sin x \cos^{2} x + {3 \sin}^{2} x \cos^{2} x + {4 \sin}^{5} x \cos^{2} x + {2 \sin}^{3} x) d x =

\cos^{2} x \sin^{2} x (3 + {4 \sin}^{3} x) + (\sin x \cos^{2} x + {2 \sin}^{3} x) =

\frac{1}{4} {(\sin 2 x)}^{2} (3 + {4 \sin}^{3} x) + \sin x (1 + \sin^{2} x) =

\frac{3}{4} \sin^{2} (2 x) + {2 \sin}^{3} x + \sin x

I = \frac{3}{4} \int \sin^{2} (2 x) d x + 3 \int \sin^{3} x d x + \int \sin x d x

\int \sin^{2} (2 x) d x u^{'} = 1 \Rightarrow u = x v = \sin^{2} (2 x) \Rightarrow v^{'} = 2 \sin (2 x) \cos (2 x) = \sin 4 x

\int \sin^{2} (2 x) d x = [x \sin^{2} (2 x)] - (\int x \sin 4 x d x = [- \frac{1}{4} x \cos 4 x] + \sin 4 x)

\Rightarrow \int \sin^{2} 2 x d x = x \sin^{2} 2 x + \frac{1}{4} (x \cos 4 x) - \sin 4 x

\int (\sin^{3} x + \sin x) d x = \int \sin x (1 + \sin^{2} x) = - \cos x (1 + \sin^{2} x) - \frac{\cos 2 x}{2}

I = \frac{3}{4} {[x \sin^{2} 2 x]}_{\frac{π}{4}}^{\frac{15 π}{4}} + \frac{3}{16} {[x \cos 4 x]}_{\frac{π}{4}}^{\frac{15 π}{4}} - \frac{3}{4} {[\sin 4 x]}_{\frac{π}{4}}^{\frac{15 π}{4}}

- {[\cos x (1 + \sin^{2} x)]}_{\frac{π}{4}}^{\frac{15 π}{4}} + {[\frac{\cos 2 x}{2}]}_{\frac{π}{4}}^{\frac{15 π}{4}}

= \frac{3}{4} [(\frac{15 π}{4} \sin^{2} (- \frac{π}{4}) - \frac{π}{4} \sin^{2} \frac{π}{4})] + [\frac{3}{16} (\frac{15 π}{4} \cos (- π) - \frac{π}{4} \cos π)] - \frac{3}{4} (\sin 15 π - \sin π)

- [\cos (\frac{π}{4}) (1 + \sin^{2} \frac{π}{4}) - (\cos \frac{π}{4} (1 + \sin^{2} \frac{π}{4})] + (\frac{1}{2} \cos \frac{3 π}{2} - \frac{1}{2} \cos \frac{π}{2})

I = \frac{- 49 π}{32}