Answer
$$\int7\cos^7tdt=7\sin t-7\sin^3 t+\frac{21\sin^5t}{5}-\sin^7t+C$$
Work Step by Step
$$A=\int7\cos^7tdt$$
This is Case 2, so we need to find a way to rewrite $$(\cos^2x)^k\cos x=(1-\sin^2x)^k\cos x$$
We have, $$A=\int7(\cos^2t)^3\cos tdt$$ $$A=7\int(1-\sin^2t)^3d(\sin t)$$
We would take $u=\sin t$
$$A=7\int(1-u^2)^3du$$ $$A=7\int(1-3u^2+3u^4-u^6)du$$ $$A=7\Big(u-u^3+\frac{3u^5}{5}-\frac{u^7}{7}\Big)+C$$ $$A=7\sin t-7\sin^3 t+\frac{21\sin^5t}{5}-\sin^7t+C$$
