Find out the answer to "Evaluating ∫−π4π4(sin⁡x+2cos⁡x)3dx"

veceriraby

Answered question

2022-01-23

Evaluating

\int_{- \frac{π}{4}}^{\frac{π}{4}} {(\sin x + 2 \cos x)}^{3} d x

Damian Roberts

Beginner2022-01-24Added 14 answers

Just expand the binomial

{(\sin x + 2 \cos x)}^{3} = \sin^{3} x + 6 \sin^{2} x \cos x + 12 \sin x \cos^{2} x + 8 \cos^{3} x

Now note that

\int_{- \frac{π}{4}}^{\frac{π}{4}} \sin^{3} x d x = 0 = \int_{- \frac{π}{4}}^{\frac{π}{4}} 12 \sin x \cos^{2} x d x

(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{π}{4}}^{\frac{π}{4}} (6 \sin^{2} x \cos x + 8 \cos^{3} x) d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} (- 2 \sin^{2} x + 8) \cos x d x

and if we set

u = \sin x

, the integral is much easier :

\int_{- \frac{π}{4}}^{\frac{π}{4}} (- 2 \sin^{2} x + 8) \cos x d x = \int_{- \frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (- 2 u^{2} + 8) d u = \frac{23 \sqrt{2}}{3}

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