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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2021 · Mathematics · Indefinite Integration

What is ∫ d x ^2( ^ -1 x) equal to ?

What is ∫dxsec⁡2(tan⁡−1x)\int \frac{d x}{\sec ^2\left(\tan ^{-1} x\right)} equal to ?

Options

  1. Asin⁡−1x+c\sin ^{-1} x+c
  2. Btan⁡−1x+c\tan ^{-1} x+c
  3. Csec⁡−1x+c\sec ^{-1} x+c
  4. Dcos⁡−1x+c\cos ^{-1} x+c

Correct answer

B. tan⁡−1x+c\tan ^{-1} x+c

Step-by-step solution

\int \frac{d x}{\sec ^2\left(\tan ^{-1} x\right)} Let tan⁡−1x=y⇒tan⁡y=x\tan ^{-1} x=y \Rightarrow \tan y=x and sec⁡2y=1+tan⁡2y=1+x2∴∫dxsec⁡2(tan⁡−1x)=∫dx1+x2=tan⁡−1x+C\begin{aligned} & \sec ^2 y=1+\tan ^2 y=1+x^2 \\ & \therefore \int \frac{d x}{\sec ^2\left(\tan ^{-1} x\right)}=\int \frac{d x}{1+x^2}=\tan ^{-1} x+C\end{aligned}
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