Question
Question: \[\frac{1}{ab}\cot^{- 1}\left( \frac{b\cos x}{a} \right) + c\]...
ab1cot−1(abcosx)+c
A
ab1cot−1(bacosx)+c
B
∫secxlog(secx+tanx)6mudx=
C
[log(secx+tanx)]2+c
D
21[log(secx+tanx)]2+c
Answer
∫secxlog(secx+tanx)6mudx=
Explanation
Solution
Put ∫x2+sin2x+2xcos2x+x+16mu⥂dx=.