Integral Calculus Question 127
Question: If $ I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp=f(p)+c} $ , then f(p) is equal to:
Options:
A) $ \frac{1}{2}\ell n[ p-\sqrt{p^{2}-1} ] $
B) $ \frac{1}{2}{{\cos }^{-1}}p+\frac{1}{2}{{\sec }^{-1}}p $
C) $ \ell n\sqrt{p+\sqrt{p^{2}-1}}-\frac{1}{2}{{\sec }^{-1}}p $
D) None of the above.
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{1}{2p}\sqrt{\frac{p-1}{p+1}}dp} $ $ =\frac{1}{2}\int{\frac{p-1}{p\sqrt{(p+1)(p-1)}}dp} $ $ =\frac{1}{2}\int{\frac{pdp}{p\sqrt{p^{2}-1}}-\frac{1}{2}\int{\frac{dp}{p\sqrt{p^{2}-1}}}} $ $ =\frac{1}{2}\int{\frac{dp}{\sqrt{p^{2}-1}}-\frac{1}{2}\int{\frac{dp}{p\sqrt{p^{2}-1}}}} $ $ =\frac{1}{2}\log _{e}( p+\sqrt{p^{2}-1} )-\frac{1}{2}{{\sec }^{-1}}p $
$ \Rightarrow f(p)=\log \sqrt{p+\sqrt{p^{2}-1}}-\frac{1}{2}{{\sec }^{-1}}p $
BETA
