Let u=sinxu=sinx.
Then (du)/dx=cos xdudx=cosx, and thus du = cos x" "dxdu=cosx dx.
Substituting this into the given integral, we get
int 2^(sin x) cos x " "dx=int 2^u" "du∫2sinxcosx dx=∫2u du
color(white)(int 2^(sin x) cos x " "dx)=color(navy)(1/(ln 2))int color(navy)(ln 2) * 2^u" "du∫2sinxcosx dx=1ln2∫ln2⋅2u du
color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^u+C∫2sinxcosx dx=1ln2⋅2u+C
And since u = sin xu=sinx, we substitute back:
color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^sin x+C∫2sinxcosx dx=1ln2⋅2sinx+C
color(white)(int 2^(sin x) cos x " "dx)=2^sin x/(ln 2)+C∫2sinxcosx dx=2sinxln2+C
So int 2^(sin x) cos x " "dx=2^sinx/ln2+C∫2sinxcosx dx=2sinxln2+C.
Check:
Using the chain rule and the exponential rule for derivatives:
d/dx (a^u)=ln a * a^u*(du)/dxddx(au)=lna⋅au⋅dudx
We get
d/dx (2^(sinx)/ln 2 + C)=1/ln 2 * ln 2 * 2^sinx * cos xddx(2sinxln2+C)=1ln2⋅ln2⋅2sinx⋅cosx
color(white)(d/dx (2^(sinx)/ln 2))=2^sinx * cos xddx(2sinxln2)=2sinx⋅cosx,
which matches our integrand from above.
