Chapter 5 - Review - Exercises - Page 423: 30

$\int sin~x~cos(cos~x)~dx =-sin(cos~x)+C$

$\int sin~x~cos(cos~x)~dx$ Let $~u = cos~x$ $\frac{du}{dx} = -sin~x$ $dx = -\frac{du}{sin~x}$ $\int (sin~x)(cos~u)~(-\frac{du}{sin~x})$ $=\int -cos~u~du$ $=-sin~u+C$ $=-sin(cos~x)+C$