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

$${\int }_0^{\pi /2}\frac{{\sin }^{2}4\theta }{{\sin }^2\theta }d\theta =?$$

Answered by Berbere last updated on 24/Mar/24

$${sin}^2\left(4x\right)=4{sin}^2\left(2x\right){cos}^2\left(2x\right)$$$$=16{sin}^2\left(x\right){cos}^2\left(x\right){cos}^2\left(2x\right)$$$${\int }_0^{\frac{\pi }{2}}8{cos}^2\left(x\right){cos}^2\left(2x\right)={\int }_0^{\frac{\pi }{2}}4(cos\left(2x\right)+1)(cos\left(4x\right)+1)dx$$$$={\int }_0^{\frac{\pi }{2}}4cos\left(2x\right)cos\left(4x\right)+4cis\left(2x\right)+4cos\left(4x\right)+4dx$$$$={\int }_0^{\frac{\pi }{2}}6cos\left(2x\right)+2cos\left(6x\right)+4cos\left(4x\right)+4dx$$$$=2\pi$$$$$$

Commented by Frix last updated on 24/Mar/24

$$\frac{{\sin }^{2}4x}{{\sin }^{2}x}=2\cos 6x+4\cos 4x+6\cos 2x+4$$$$\Rightarrow$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}4x}{{\sin }^{2}x}dx=$$$$={[\frac{1}{3}\sin 6x+\sin 4x+3\sin 2x+4x]}_0^{\frac{\pi }{2}}=2\pi$$

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