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

Question

Charles Arocha · Accepted Answer

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

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#. 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.

Scarlett Allison · Answer

undefined To find the indefinite integral of $ \int 2^{\sin(x)} \cos(x) \, dx $, you would use the technique of substitution. Let $ u = \sin(x) $. Then, $ du = \cos(x) \, dx $.

Substituting $ u = \sin(x) $ and $ du = \cos(x) \, dx $, the integral becomes:

$$ \int 2^u \, du $$

This is now a standard exponential function, which integrates to:

$$ \frac{2^u}{\ln(2)} + C $$

Where $ C $ is the constant of integration.

Substituting back for $ u $, we get:

$$ \frac{2^{\sin(x)}}{\ln(2)} + C $$

So, the indefinite integral of $ \int 2^{\sin(x)} \cos(x) \, dx $ is $ \frac{2^{\sin(x)}}{\ln(2)} + C $.