Question

Using beta functions evaluate ${\displaystyle {\int }_0^{\frac{\pi }{6}} {\cos }^{6}3\theta .\sin \theta d\theta }$ 

Answer 1

let I= ${\displaystyle {\int }_0^{\frac{\pi }{6}} {\cos }^{6}3\theta .{\sin }^{2}6\theta d\theta }$ 

Put  ${\displaystyle 3\theta =t}$ 

Diff. w.r.t 

${\displaystyle d\theta =\frac{dt}{3}}$                   Limit:${\displaystyle [0,\frac{\pi }{2}]}$ 

∴ ${\displaystyle I=\frac{1}{3}\frac{{\int }_0^{\pi }}{2} {\cos }^{6}t.{\sin }^{2}2tdt}$ 

=${\displaystyle \frac{4}{3}{\int }_0^{\frac{\pi }{2}}{\cos }^{3}t{(\sin t.\cos t)}^{2}dt}$ 

=${\displaystyle \frac{4}{3}{\int }_0^{\frac{\pi }{2}}{\cos }^{5}t.{\sin }^{2}t.dt}$ 

=${\displaystyle \frac{4}{3}\times \frac{1}{2}\times \beta (3,\frac{3}{2})}$       ... ${\displaystyle \{{\int }_0^{\frac{\pi }{2}}{\cos }^{m}t.{\sin }^{n}t.=\frac{1}{2}\times \beta (m+1,n+1)\}}$ 

∴ ${\displaystyle I=\frac{32}{315}}$

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