Integral Calculus Question 425
Question: $ \int_{{}}^{{}}{\frac{d\theta }{\sin \theta {{\cos }^{3}}\theta }=} $
Options:
A) $ \log \tan \theta +{{\tan }^{2}}\theta +c $
B) $ \log \tan \theta -\frac{1}{2}{{\tan }^{2}}\theta +c $
C) $ \log \tan \theta +\frac{1}{2}{{\tan }^{2}}\theta +c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{d\theta }{\sin \theta {{\cos }^{3}}\theta }=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta ,d\theta }{\sin \theta \cos \theta }=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta (1+{{\tan }^{2}}\theta )}{\tan \theta }}}}d\theta $ Put $ t=\tan \theta \Rightarrow dt={{\sec }^{2}}\theta ,d\theta , $ then it reduces to $ \int_{{}}^{{}}{\frac{1+t^{2}}{t},dt=\int_{{}}^{{}}{( \frac{1}{t}+t ),dt}} $ $ =\log t+\frac{t^{2}}{2}+c=\log \tan \theta +\frac{{{\tan }^{2}}\theta }{2}+c. $
BETA
