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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Definite Integration

NDA 2023 · Mathematics · Definite Integration

Let I=∫_ -2 π ^ 2 π ( ^4 x+ ^4 x)/(1+3^x) d x What is ∫_0^π( ^4 x+...

Let I=∫−2π2πsin⁡4x+cos⁡4x1+3xdxI=\int_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d x What is ∫0π(sin⁡4x+cos⁡4x)dx\int_0^\pi\left(\sin ^4 x+\cos ^4 x\right) d x equal to?

Options

  1. A3π8\frac{3 \pi}{8}
  2. B3π4\frac{3 \pi}{4}
  3. C3π2\frac{3 \pi}{2}
  4. D3π3 \pi

Correct answer

B. 3π4\frac{3 \pi}{4}

Step-by-step solution

 Since, ∫0π(sin⁡4x+cos⁡4x)dx=∫0π((sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x)dx=∫0π(1−sin⁡22x2)dx=∫0π(1−1−cos⁡4x4)dx=14∫0π(3+cos⁡4x)dx=14(3x+sin⁡4x4)0π=3π4\begin{aligned} & \text{ Since, } \int_0^\pi\left(\sin ^4 x+\cos ^4 x\right) d x \\ & =\int_0^\pi\left(\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x\right) d x \\ & =\int_0^\pi\left(1-\frac{\sin ^2 2 x}{2}\right) d x=\int_0^\pi\left(1-\frac{1-\cos 4 x}{4}\right) d x \\ & =\frac{1}{4} \int_0^\pi(3+\cos 4 x) d x=\frac{1}{4}\left(3 x+\frac{\sin 4 x}{4}\right)_0^\pi=\frac{3 \pi}{4}\end{aligned}
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