Integral Calculus Question 410
Question: $ \int_{{}}^{{}}{\tan (3x-5)\sec (3x-5)\ dx=} $
[MP PET 1988]
Options:
A) $ \sec (3x-5)+c $
B) $ \frac{1}{3}\sec (3x-5)+c $
C) $ \tan (3x-5)+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ t=3x-5\Rightarrow dt=3dx, $ therefore $ \int_{{}}^{{}}{\tan (3x-5),\sec (3x-5),dx}=\frac{1}{3}\int_{{}}^{{}}{\tan t.\sec t,dt} $ $ =\frac{\sec t}{3}+c=\frac{\sec (3x-5)}{3}+c. $
BETA
