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
∫ − π 4 π 4 tan 3 x d x \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \tan ^3 x d x ∫ 4 − π 4 π tan 3 x d x equal to?
Options A 3 \sqrt{3} 3 B 1 3 \frac{1}{3} 3 1 C 1 2 \frac{1}{2} 2 1 D 0
Step-by-step solution
A = ∫ − π / 4 π / 4 tan x tan 2 x d x = ∫ − π / 4 π / 4 tan x ( sec 2 x − 1 ) d x = ∫ − π / 4 π / 4 tan x ⋅ sec 2 x d x − ∫ − π / 4 π / 4 tan x d x \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} A = ∫ − π /4 π /4 tan x tan 2 x d x = ∫ − π /4 π /4 tan x ( sec 2 x − 1 ) d x = ∫ − π /4 π /4 tan x ⋅ sec 2 x d x − ∫ − π /4 π /4 tan x d x
Put
tan x = t \tan x=t tan x = t or
sec 2 x d x = d t \sec ^2 x d x=d t sec 2 x d x = d t A = ∫ − 1 1 t d t − [ log sec x ] − π / 4 π / 4 = [ t 2 2 ] − 1 1 − [ 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} A = ∫ − 1 1 t d t − [ log sec x ] − π /4 π /4 = [ 2 t 2 ] − 1 1 − [ log sec ( π /4 ) − log sec ( − π /4 )] = 0 − [ log 2 − log 2 ] = 0 $$
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