Question
Question: \[\log(a^{2} + b^{2}\sin^{2}x) + c\]...
log(a2+b2sin2x)+c
A
b2log(a2+b2sin2x)+c
B
∫x1+logx16mudx=
C
32(1+logx)3/2+c
D
(1+logx)3/2+c
Answer
b2log(a2+b2sin2x)+c
Explanation
Solution
(1−x)−1−1+c
Put ∫(cosx−sinx)⥂6mudx=2sin(x+α)+c then it reduces to
α=
3π
−3π
4π, −4πconstant).