Step by step solution to "Easiest way to prove that∫π6π2sin⁡(2x)3cos⁡(3x)2dx=(34)4"

Nettie Potts

Answered question

2022-02-26

Easiest way to prove that

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

Haiden Frazier

Beginner2022-02-27Added 10 answers

\int_{\frac{π}{6}}^{\frac{π}{2}} {\sin (2 x)}^{3} {\cos (3 x)}^{2} d x

Since

{(\sin 2 x)}^{2} = 1 - \cos 4 x

and

{(\cos 3 x)}^{2} = 1 + \cos 6 x

= \frac{1}{4} \int_{\frac{π}{6}}^{\frac{π}{2}} (\sin 2 x) (1 - \cos 4 x) (1 + \cos 6 x)

Substitute

u = 2 x

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

Use the formula

\sin u = \frac{e^{t u} - e^{- t u}}{2 i}

and

\cos u = \frac{e^{t u} + e^{- t u}}{2}

= \frac{1}{8} \int_{\frac{π}{3}}^{π} \frac{e^{i u} - e^{- i u}}{2 i} (1 - \frac{e^{2 i u} + e^{- 2 i u}}{2}) (1 + \frac{e^{3 i u} + e^{- 3 i u}}{2}) d u

Expand,

= \frac{1}{64 i} \int_{\frac{π}{3}}^{π} (- (e^{6 i u} - e^{- 6 i u}) + 3 (e^{4 i u} - e^{- 4 i u}) - 2 (e^{3 i u} - e^{- 3 i u}) - 3 (e^{2 i u} - e^{- 2 i u}) + 6 (e^{i u} - e^{- i u})) d u

Use the formula

\sin u = \frac{e^{t u} - e^{- t u}}{2 i}

, go back to

\sin

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