Integral of sin(2*t) dx
The solution
Detail solution
-
There are multiple ways to do this integral.
Method #1
-
Let u=2t.
Then let du=2dt and substitute 2du:
∫2sin(u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫sin(u)du=2∫sin(u)du
-
The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −2cos(u)
Now substitute u back in:
−2cos(2t)
Method #2
-
The integral of a constant times a function is the constant times the integral of the function:
∫2sin(t)cos(t)dt=2∫sin(t)cos(t)dt
-
There are multiple ways to do this integral.
Method #1
-
Let u=cos(t).
Then let du=−sin(t)dt and substitute −du:
∫(−u)du
-
The integral of a constant times a function is the constant times the integral of the function:
∫udu=−∫udu
-
The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −2u2
Now substitute u back in:
−2cos2(t)
Method #2
-
Let u=sin(t).
Then let du=cos(t)dt and substitute du:
-
The integral of un is n+1un+1 when n=−1:
∫udu=2u2
Now substitute u back in:
2sin2(t)
So, the result is: −cos2(t)
-
Add the constant of integration:
−2cos(2t)+constant
The answer is:
−2cos(2t)+constant
The answer (Indefinite)
[src]
/
| cos(2*t)
| sin(2*t) dt = C - --------
| 2
/
∫sin(2t)dt=C−2cos(2t)
The graph
21−2cos(2)
=
21−2cos(2)
Use the examples entering the upper and lower limits of integration.