Perform the following indefinite integrals ∫Cos(10x) dx?

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Answer 1 · 2018-03-25T19:14:17

#intcos(10x)dx=1/10sin(10x)+C#

In general,

#intcos(ax)dx=1/asin(ax)+C#. I'll show why at the end.

Thus,

#intcos(10x)dx=1/10sin(10x)+C#

Why? We could make the substitution

#u=10x#

#(du)/dx=10#

#du=10dx#

#dx=1/10du#

Thus,

#intcos(10x)dx=1/10intcos(u)du=1/10sin(u)+C=1/10sin(10x)+C#

Answer 2 · 2018-03-25T19:19:39

#=> 1/10sin(10x)+C#

where #C# is an arbitrary constant from indefinite integration.

#=>\int cos(10x)dx#

Let #u equiv 10x#. This means #du = 10dx -> dx = (du)/10#.

Now let's make the substitution:

#=>\int cos(u)(du)/10#

#=>1/10\int cos(u)du#

Now we can integrate:

#=>1/10sin(u)+C#

Substituting back in for #u#:

#=> 1/10sin(10x)+C#

where #C# is an arbitrary constant from indefinite integration.