Question
Mathematics Question on Integrals of Some Particular Functions
∫1+cosxdx is equal to
A
22cos2x+C
B
22sin2x+C
C
2cos2x+C
D
2sin2x+C
Answer
22sin2x+C
Explanation
Solution
∫1+cosxdx=2∫cos(2x)dx
=22sin(2x)+c
[∵1+cosx=2cos22x]
∫1+cosxdx is equal to
22cos2x+C
22sin2x+C
2cos2x+C
2sin2x+C
22sin2x+C
∫1+cosxdx=2∫cos(2x)dx
=22sin(2x)+c
[∵1+cosx=2cos22x]