∫(from 0 to pi/2) 7^cost sint dt

i know that it is a^u du

but sin needs to be negative so I tried this:

-∫(from 0 to pi/2) 7^cost (-sint) dt

then made u=cost and du= -sint

so:∫7^u du= 7^u/ ln7= 7^cost /ln7

and when i plug in pi/2 i get 1/ln7

but the answer is 6/ln7