Question
Question: \[\text{cosec}x\cot x + c\]...
cosecxcotx+c
A
∫{1+2tanx(tanx+secx)}1/2dx=
B
log(secx+tanx)+c
C
log(secx+tanx)1/2+c
D
None of these
Answer
log(secx+tanx)+c
Explanation
Solution
logtanθ+21tan2θ+c
∫cos−1x.1−x21dx=.
cosecxcotx+c
∫{1+2tanx(tanx+secx)}1/2dx=
log(secx+tanx)+c
log(secx+tanx)1/2+c
None of these
log(secx+tanx)+c
logtanθ+21tan2θ+c
∫cos−1x.1−x21dx=.