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Question Number 205558 by universe last updated on 24/Mar/24

\int_{0}^{π / 2} \frac{\sin^{2} 4 θ}{\sin^{2} θ} d θ = ?

Answered by Berbere last updated on 24/Mar/24

{s i n}^{2} (4 x) = 4 {s i n}^{2} (2 x) {c o s}^{2} (2 x)

= 16 {s i n}^{2} (x) {c o s}^{2} (x) {c o s}^{2} (2 x)

\int_{0}^{\frac{π}{2}} 8 {c o s}^{2} (x) {c o s}^{2} (2 x) = \int_{0}^{\frac{π}{2}} 4 (c o s (2 x) + 1) (c o s (4 x) + 1) d x

= \int_{0}^{\frac{π}{2}} 4 c o s (2 x) c o s (4 x) + 4 c i s (2 x) + 4 c o s (4 x) + 4 d x

= \int_{0}^{\frac{π}{2}} 6 c o s (2 x) + 2 c o s (6 x) + 4 c o s (4 x) + 4 d x

= 2 π

Commented by Frix last updated on 24/Mar/24

\frac{\sin^{2} 4 x}{\sin^{2} x} = 2 \cos 6 x + 4 \cos 4 x + 6 \cos 2 x + 4

\Rightarrow

\underset{0}{\int^{\frac{π}{2}}} \frac{\sin^{2} 4 x}{\sin^{2} x} d x =

= {[\frac{1}{3} \sin 6 x + \sin 4 x + 3 \sin 2 x + 4 x]}_{0}^{\frac{π}{2}} = 2 π

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