How do you integrate #int 2^sinxcosxdx#?

Question

1791 views around the world

You can reuse this answer
Creative Commons License

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#