Use-gamma-function-to-prove-i-0-8-cos-3-4x-dx-1-6-ii-0-6-cos-4-3-sin-2-6-d-5-192-

Question Number 90574 by niroj last updated on 24/Apr/20

\boldsymbol Use \boldsymbol gamma \boldsymbol function \boldsymbol to \boldsymbol prove

(i) . \int_{0}^{\frac{\boldsymbol π}{8}} \boldsymbol \cos^{3} 4 \boldsymbol x \boldsymbol dx = \frac{1}{6} .

(\boldsymbol ii) . \int_{0}^{\frac{\boldsymbol π}{6}} \boldsymbol \cos^{4} 3 \boldsymbol θ \boldsymbol \sin^{2} 6 \boldsymbol θ \boldsymbol d θ = \frac{5 \boldsymbol π}{192} .

Commented by mathmax by abdo last updated on 24/Apr/20

I = \int_{0}^{\frac{π}{8}} {c o s}^{3} (4 x) d x \Rightarrow I =_{4 x = t} \frac{1}{4} \int_{0}^{\frac{π}{2}} {c o s}^{3} (t) d t

= \frac{1}{4} \int_{0}^{\frac{π}{2}} {c o s}^{2} t s i n t d t

B (p, q) = \int_{0}^{\frac{π}{2}} {c o s}^{2 p - 1} x {s i n}^{2 q - 1} x d x

\Rightarrow \int_{0}^{\frac{π}{2}} {c o s}^{2} t {s i n}^{2} t d t = \int_{0}^{\frac{π}{2}} {c o s}^{2 (\frac{3}{2}) - 1} t {s i n}^{2 (1) - 1} t d t

= B (\frac{3}{2}, 1) = \frac{Γ (\frac{3}{2}) . Γ (1)}{Γ (\frac{3}{2} + 1)} w e h a v e Γ (n) = (n - 1)! \Rightarrow Γ (1) = 1

Γ (x + 1) = x Γ (x) \Rightarrow

Γ (\frac{3}{2} + 1) = \frac{3}{2} Γ (\frac{3}{2}) \Rightarrow B (\frac{3}{2}, 1) = \frac{Γ (\frac{3}{2})}{\frac{3}{2} Γ (\frac{3}{2})} = \frac{2}{3} \Rightarrow

I = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}

Commented by niroj last updated on 25/Apr/20

thank you .

thank you .

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