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

Substituting this into the given integral, we get

We get

which matches our integrand from above.

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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 ).

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