What is the integral of #cos2(theta)#?

1 Answer
Mar 13, 2016

#int cos(2theta) "d"theta = 1/2 sin(2theta) + C#,

where #C# is an integration constant.

Explanation:

I think you mean #cos(2theta)# instead of #cos2(theta)#.

If you know that #int cos(x) dx = sin(x) + C#, then we can use a substitution (which is the reverse of the chain rule).

Let #u = 2theta#,

#frac{"d"u}{"d"theta} = 2#.

So,

#int cos(2theta) "d"theta = 1/2 int cos(2theta) * (2) "d"theta#

#= 1/2 int cos(2theta) * frac{"d"u}{"d"theta} "d"theta#

#= 1/2 int cos(u) "d"u#

#= 1/2 (sin(u) + C_1)#,
where #C_1# is an integration constant.

#= 1/2 sin(2theta) + C_2#,
where #C_2 = 1/2 C_1#.