Question
Question: \[\log(1 + \cot x) + c\]...
log(1+cotx)+c
A
−log(1+cotx)+c
B
2(1+cotx)21+c
C
∫x1sinx6mudx=
D
−21cosx+c
Answer
∫x1sinx6mudx=
Explanation
Solution
logcotx+c
Put ∫1+sinx1dx= then it reduces to
22logtan(8π+4x)+c.
log(1+cotx)+c
−log(1+cotx)+c
2(1+cotx)21+c
∫x1sinx6mudx=
−21cosx+c
∫x1sinx6mudx=
logcotx+c
Put ∫1+sinx1dx= then it reduces to
22logtan(8π+4x)+c.