What is the integral of #10sin(x/5)#?

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Answer 1 · 2015-04-22T14:53:18

It is #-50cos(x/5)#

#int 10 sin(x/5) dx = 10 intsin(x/5) dx#

Let #u = x/5# so that #du = 1/5 dx# and therefore #dx = 5 du#

Substitute:

#10 int sin(x/5) dx = 10 int (sinu) 5 du = 50 int sinu du#

Now integrate:

#10 int sin(x/5) dx = 50 (-cos u) +C
And finally, reverse the substitution:

#10 int sin(x/5) dx = -50 cos (x/5)+C#