How do you find the indefinite integral of #int 2^(sinx)cosx#?

1 Answer
Dec 24, 2016

Answer:

#int 2^(sin x) cos x " "dx=2^sinx/ln2+C#.

Explanation:

Let #u=sinx#.
Then #(du)/dx=cos x#, and thus #du = cos x" "dx#.

Substituting this into the given integral, we get

#int 2^(sin x) cos x " "dx=int 2^u" "du#
#color(white)(int 2^(sin x) cos x " "dx)=color(navy)(1/(ln 2))int color(navy)(ln 2) * 2^u" "du#
#color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^u+C#

And since #u = sin x#, we substitute back:

#color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^sin x+C#
#color(white)(int 2^(sin x) cos x " "dx)=2^sin x/(ln 2)+C#

So #int 2^(sin x) cos x " "dx=2^sinx/ln2+C#.

Check:

Using the chain rule and the exponential rule for derivatives:
#d/dx (a^u)=ln a * a^u*(du)/dx#

We get

#d/dx (2^(sinx)/ln 2 + C)=1/ln 2 * ln 2 * 2^sinx * cos x#
#color(white)(d/dx (2^(sinx)/ln 2))=2^sinx * cos x#,

which matches our integrand from above.