Question
Question: \[{logsin}x + c\]...
logsinx+c
A
logsecx+c
B
∫x416mudx
C
−3x31+c
D
3x31+c
Answer
−3x31+c
Explanation
Solution
−2cosx+c
21cosx+c2cosx+c
⇒I=−∫tan2xdx tan(ex)−x+c
⇒I=−∫sec2xdx+∫1dx sec(ex)+c
⇒ tan(ex)−ex+c
logsinx+c
logsecx+c
∫x416mudx
−3x31+c
3x31+c
−3x31+c
−2cosx+c
21cosx+c2cosx+c
⇒I=−∫tan2xdx tan(ex)−x+c
⇒I=−∫sec2xdx+∫1dx sec(ex)+c
⇒ tan(ex)−ex+c