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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Definite Integration

NDA 2022 · Mathematics · Definite Integration

If ∫_0^ π / 2 ( ^4 x+ ^4 x) d x=k , then what is the value of ∫_0^...

If ∫0π/2(sin⁡4x+cos⁡4x)dx=k\int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x=k, then what is the value of ∫020π(sin⁡4+cos⁡4x)dx?\int_0^{20 \pi}\left(\sin ^4+\cos ^4 x\right) d x ?

Options

  1. Akk
  2. B10k10 k
  3. C20k20 k
  4. D40k40 k

Correct answer

D. 40k40 k

Step-by-step solution

Concept: If f(x)\mathrm{f}(\mathrm{x}) is a periodic function with period T , then ∫0nTf(x)dx=n∫0Tf(x)dx\int_0^{n T} f(x) d x=n \int_0^T f(x) d x If f(x)f(x) is periodic with period TT, then f(ax)f(a x) is periodic with period T∣a∣\frac{T}{|a|} Calculation:  Let f(x)=(sin⁡4x+cos⁡4x)⇒f(x)=(sin⁡2x)2+(cos⁡2x)2+2sin⁡2xcos⁡2x−2sin⁡2xcos⁡2x⇒f(x)=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x⇒f(x)=1−sin⁡22x2⇒f(x)=1−1−cos⁡4x4⇒f(x)=34+cos⁡4x4\begin{aligned} & \text{ Let } f(x)=\left(\sin ^4 x+\cos ^4 x\right) \\ & \Rightarrow f(x)=\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^2+2 \sin ^2 x \cos ^2 x-2 \sin ^2 x \cos ^2 x \\ & \Rightarrow f(x)=\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x \\ & \Rightarrow f(x)=1-\frac{\sin ^2 2 x}{2} \\ & \Rightarrow f(x)=1-\frac{1-\cos 4 x}{4} \\ & \Rightarrow f(x)=\frac{3}{4}+\frac{\cos 4 x}{4} \end{aligned} Since cos⁡x\cos x is a function of period 2π⇒cos⁡4x2 \pi\Rightarrow \cos 4 x is periodic with period 2π4=π2⇒f(x)=(sin⁡4x+cos⁡4x)\frac{2 \pi}{4}=\frac{\pi}{2}\Rightarrow f(x)= \left(\sin ^4 x+\cos ^4 x\right) is periodic with period π2\frac{\pi}{2} Given, ∫0π/2(sin⁡4x+cos⁡4x)dx=k\int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x=k\qquad So, ∫020π(sin⁡4x+cos⁡4x)dx=∫040(π2)(sin⁡4x+cos⁡4x)dx\int_0^{20 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=\int_0^{40 \left(\frac{\pi}{2}\right)}\left(\sin ^4 x+\cos ^4 x\right) d x As f(x)=(sin⁡4x+cos⁡4x)f(x)=\left(\sin ^4 x+\cos ^4 x\right) is periodic with period π2\frac{\pi}{2} By the property of definite integral, ∫040(π2)(sin⁡4x+cos⁡4x)dx=40∫0π/2(sin⁡4x+cos⁡4x)dx⇒∫020π(sin⁡4x+cos⁡4x)dx=40k{ Using (i)}\begin{aligned} & \int_0^{40\left(\frac{\pi}{2}\right)}\left(\sin ^4 x+\cos ^4 x\right) d x=40 \int_0^{\pi / 2}\left(\sin ^4 x+\cos ^4 x\right) d x \\ & \Rightarrow \int_0^{20 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=40 k\{\text{ Using }(i)\} \end{aligned}
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