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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Definite Integration

NDA 2010 · Mathematics · Definite Integration

What is ∫_ (-π)/(4) ^ (π)/(4) ^3 x d x equal to?

What is ∫−π4π4tan⁡3xdx\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \tan ^3 x d x equal to?

Options

  1. A3\sqrt{3}
  2. B13\frac{1}{3}
  3. C12\frac{1}{2}
  4. D0

Correct answer

D. 0

Step-by-step solution

A=∫−π/4π/4tan⁡xtan⁡2xdx=∫−π/4π/4tan⁡x(sec⁡2x−1)dx=∫−π/4π/4tan⁡x⋅sec⁡2xdx−∫−π/4π/4tan⁡xdx\begin{aligned} \mathrm{A} & =\int_{-\pi / 4}^{\pi / 4} \tan x \tan ^2 x d x \\ & =\int_{-\pi / 4}^{\pi / 4} \tan x\left(\sec ^2 x-1\right) d x \\ & =\int_{-\pi / 4}^{\pi / 4} \tan x \cdot \sec ^2 x d x-\int_{-\pi / 4}^{\pi / 4} \tan x d x \end{aligned} Put tan⁡x=t\tan x=t or sec⁡2xdx=dt\sec ^2 x d x=d tA=∫−11tdt−[log⁡sec⁡x]−π/4π/4=[t22]−11−[log⁡sec⁡(π/4)−log⁡sec⁡(−π/4)]=0−[log⁡2−log⁡2]=0\begin{aligned} & A=\int_{-1}^1 t d t-[\log \sec x]_{-\pi / 4}^{\pi / 4} \\ & =\left[\frac{t^2}{2}\right]_{-1}^1-[\log \sec (\pi / 4)-\log \sec (-\pi / 4)] \\ & =0-[\log \sqrt{2}-\log \sqrt{2}]=0\end{aligned}$$
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