integrate sinx/1+cosx dx
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Answered by
1
Log (1/1+cosx) is the anti derivative of this function..
Hope it helps..
Hope it helps..
Answered by
0
Hello!
Your Answer:
It
Problem:
∫sin(x)cos(x)+1dx
Substitute u=cos(x)+1
⟶ dx=−1sin(x)du
(Steps):
=−∫1udu
Now solving:
∫1udu
This is a standard integral:
=ln(u)
Plug in solved integrals:
−∫1udu
=−ln(u)
Undo substitution u=cos(x)+1
:
=−ln(cos(x)+1)
The problem is solved:
∫sin(x)cos(x)+1dx
=−ln(cos(x)+1)+C
Hope It Helps.
Have A Nice Day.
Your Answer:
It
Problem:
∫sin(x)cos(x)+1dx
Substitute u=cos(x)+1
⟶ dx=−1sin(x)du
(Steps):
=−∫1udu
Now solving:
∫1udu
This is a standard integral:
=ln(u)
Plug in solved integrals:
−∫1udu
=−ln(u)
Undo substitution u=cos(x)+1
:
=−ln(cos(x)+1)
The problem is solved:
∫sin(x)cos(x)+1dx
=−ln(cos(x)+1)+C
Hope It Helps.
Have A Nice Day.
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