#int [(sin t)^7][(cos t)^4]dt#
We know that #d/(dx)(sinx)=cosx# and #d/(dx)(cosx)= - sinx#.
We'd like to do a substitution. We also know that :
#(sint)^2+(cost)^2=1#
Those are the key ideas.
We have #sint# raised to an odd power and #cost# raised to a power, so if we group one #sint# witlh #dt# and change the remaining even power of #sint# to an expression involving only #cost#, we will be able to substitute.
#int [(sin t)^7][(cos t)^4]dt = int [(sin t)^2]^3[(cos t)^4][sint]dt #
#= int [1-(cos t)^2]^3 [(cos t)^4][sint]dt #
Let #u=cost#, then the integral becomes:
#= - int [1-u^2]^3 [u^4]du #
Expand the polynomial and integrate, then back-substitute for #u# back to #cost#.
