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Question 75 of 96

Q.The value of ∫0πsin⁡4x dx\displaystyle\int_{0}^{\pi} \sin^4 x\, dx is :

(a) 3π2\dfrac{3\pi}{2}
(b) 3π10\dfrac{3\pi}{10}
(c) 3π8\dfrac{3\pi}{8}
(d) 3π4\dfrac{3\pi}{4}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2020MCQ· 1mImportance★★★★★
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Using the symmetry of sin⁡4x\sin^4x about x=π/2x=\pi/2 and Wallis' reduction formula, ∫0πsin⁡4x dx=3π8\displaystyle\int_0^{\pi}\sin^4x\,dx=\dfrac{3\pi}{8}.

  1. Since sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x, the graph of sin⁡4x\sin^4x on [0,π][0,\pi] is symmetric about x=π/2x=\pi/2, so ∫0πsin⁡4x dx=2∫0π/2sin⁡4x dx\displaystyle\int_0^{\pi}\sin^4x\,dx = 2\int_0^{\pi/2}\sin^4x\,dx.
  2. By Wallis' (reduction) formula, for even nn, ∫0π/2sin⁡nx dx=(n−1)(n−3)⋯1n(n−2)⋯2⋅π2\displaystyle\int_0^{\pi/2}\sin^n x\,dx=\dfrac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\dfrac{\pi}{2}. …

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