0-pi-2-sin-2-x-cos-6-x-dx-

Question Number 139285 by mohammad17 last updated on 25/Apr/21

\int_{0}^{\frac{π}{2}} {s i n}^{2} (x) {c o s}^{6} (x) d x

Answered by qaz last updated on 25/Apr/21

\int_{0}^{π / 2} \sin^{2} x \cos^{6} x d x = \frac{1}{2} B (\frac{3}{2}, \frac{7}{2})

= \frac{\frac{1 \times 5 \times 3 \times 1}{2 \times 2 \times 2 \times 2} π}{2 \times 4 \times 3 \times 2} = \frac{5 π}{256}

Commented by mohammad17 last updated on 25/Apr/21

t h a n k y o u s i r c a n y o u g i v e m e s t e b b y s t e b

p l e a s e ?

Commented by qaz last updated on 25/Apr/21

\int_{0}^{π / 2} \sin^{p} x \cos^{q} x d x = \frac{1}{2} B (\frac{p + 1}{2}, \frac{q + 1}{2})

= \frac{Γ (\frac{p + 1}{2}) Γ (\frac{q + 1}{2})}{2 Γ (\frac{p + q + 2}{2})}

Γ (x + 1) = x Γ (x)

Γ (\frac{1}{2}) = \sqrt{π}

Answered by MJS_new last updated on 25/Apr/21

\int \sin^{2} x \cos^{6} x d x =

[t = \tan x \to d x = \cos^{2} x d t]

= \int \frac{t^{2}}{{(t^{2} + 1)}^{5}} d t =

[{Ostrogradski}^{'} s Method]

= \frac{t (15 t^{6} + 55 t^{4} + 73 t^{2} - 15)}{384 {(t^{2} + 1)}^{4}} + \frac{5}{128} \int \frac{d t}{t^{2} + 1} =

= \frac{t (15 t^{6} + 55 t^{4} + 73 t^{2} - 15)}{384 {(t^{2} + 1)}^{4}} + \frac{5}{128} \arctan t

\Rightarrow answer is \frac{5 π}{256}

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