Question

$\int \frac{d(\cos \theta )}{\sqrt{1-{\cos }^2\theta }}=$

• A. ${\cos }^{-1}\theta +C$
• B. $\theta +C$
• C. ${\sin }^{-1}\theta +C$
• D. ${\sin }^{-1}(\cos \theta )+C$

Answer 1

The correct option is D

${\sin }^{-1}(\cos \theta )+C$

Explanation for the correct option:

Evaluating the integral:

$\begin{array}{rcl}I& =& \int \frac{d(\cos \theta )}{\sqrt{1–{\cos }^2\theta }}\\ & =& –\int \frac{\sin \theta d\theta }{\sqrt{1–{\cos }^2\theta }}\\ \text{Let}t& =& \cos \theta \\ & \Rightarrow & dt=-\sin \theta d\theta \\ I& =& \int \frac{dt}{\sqrt{1–t^2}}\left[\text{substituting}t\text{and}dt\right]\\ & =& {\sin }^{-1}t+C\\ & =& {\sin }^{-1}(\cos \theta )+C\end{array}$

Therefore, $\int \frac{d(\cos \theta )}{\sqrt{1-{\cos }^2\theta }}=$${\sin }^{-1}(\cos \theta )+C$

Hence the correct answer is option (D).