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.
