Integral Calculus Question 469
Question: $ \int{\frac{{{\sin }^{3}}2x}{{{\cos }^{5}}2x}dx=} $
[Karnataka CET 1999]
Options:
A) $ {{\tan }^{4}}x+C $
B) $ \tan 4x+C $
C) $ {{\tan }^{4}}2x+x+C $
D) $ \frac{1}{8}{{\tan }^{4}}2x+C $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int{\frac{{{\sin }^{3}}2x}{{{\cos }^{5}}2x}dx} $
Þ $ I=\int{\frac{{{\sin }^{3}}2x}{{{\cos }^{3}}2x}.\frac{1}{{{\cos }^{2}}2x}dx=\int{{{\tan }^{3}}2x.{{\sec }^{2}}2x,dx.}} $ Putting tan $ 2x=t $ and $ 2{{\sec }^{2}}2x,dx=dt $ , we get $ I=\int{t^{3}\frac{dt}{2}=\frac{1}{2}.\frac{t^{4}}{4}+C=\frac{1}{8}({{\tan }^{4}}2x)+C.} $
BETA
