How do you prove #intcosAxdx = 1/AsinAx+c# ?

1 Answer
maganbhai P.
May 5, 2018

Please see below.

Explanation:

We know that,

#(1)d/(dx)(sinx)=cosx#

#=>color(red)(intcosxdx=sinx+c# ( definition of antiderivative )

Let

#I=intcosAxdx#

We take,

#Ax=u=>Adx=du=>dx=1/Adu#

So,

#I=intcosu*1/Adu#

#=>I=1/Aintcosudu...tocolor(red)(Apply(1)#

#=>I=1/Asinu+c,where,u=Ax#

#=>I=1/AsinAx+c#

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