How do you integrate $int 2^sinxcosxdx$ ?

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Answer 1 · 2016-12-06T15:42:18

The answer is $=2^(sinx)/ln2+C$

We do this by substitution

Let, $sinx=u$

Then, $cosxdx=du$

Integral, $I=int2^(sinx)cosxdx$

$=int2^udu$

Let $y=2^u$

$lny=u ln2$

$y=e^(u ln2)=2^u$

Therefore,

$I=inte^(u ln2)du=e^(u ln2)/ln2=2^u/ln2=2^(sinx)/ln2+C$

How do you integrate int 2^sinxcosxdxint 2^sinxcosxdx?