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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 I equal to?

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 I equal to?

Options

  1. A0
  2. B3π4\frac{3 \pi}{4}
  3. C3π2\frac{3 \pi}{2}
  4. D3π3 \pi

Correct answer

C. 3π2\frac{3 \pi}{2}

Step-by-step solution

Since, I=∫−2π2πsin⁡4x+cos⁡4x1+3xdxI=∫−2π2π3x(sin⁡4x+cos⁡4x)1+3xdxI=\int_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d xI=\int_{-2 \pi}^{2 \pi} \frac{3^x\left(\sin ^4 x+\cos ^4 x\right)}{1+3^x} d x Since, 2I=∫−2π2π(sin⁡4x+cos⁡4x)dx⇒2I=14(3x+sin⁡4x4)−2π2π⇒22 I=\int_{-2 \pi}^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x\Rightarrow 2 I=\frac{1}{4}\left(3 x+\frac{\sin 4 x}{4}\right)_{-2 \pi}^{2 \pi}\Rightarrow 2\mathrm{I}=14(3x(4π))⇒=\frac{1}{4}(3 \mathrm{x}(4 \pi)) \Rightarrow\mathrm{I}=3π2=\frac{3 \pi}{2}
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