Answer
\begin{aligned}
\int 7 \cos ^{7} t \ dt =7 \sin t-7 \sin ^{3} t+\frac{21}{5} \sin ^{5} t-\sin ^{7} t+C
\end{aligned}
Work Step by Step
Given $$ \int 7 \cos ^{7}t\ dt $$
So, we have
\begin{aligned}
I&= \int 7 \cos ^{7} t dt \\
&= \int 7 \cos t\ (\cos ^{2} t)^3 dt \\
&= \int 7 \cos t\ (1-\sin ^{2} t)^3 dt \\
&= \int 7 \cos t\ (1-3\sin^2t+3\sin^4t-\sin ^{6} t) dt \\
&=7\left[\int \cos t\ d t-3 \int \sin ^{2} t \cos t\ d t\\
+3 \int \sin ^{4} t \cos t\ d t-\int \sin ^{6} t \cos t \ d t\right]\\
&=7\left(\sin t-3 \frac{\sin ^{3} t}{3}+3 \frac{\sin ^{5} t}{5}-\frac{\sin ^{7} t}{7}\right)+C\\
&=7 \sin t-7 \sin ^{3} t+\frac{21}{5} \sin ^{5} t-\sin ^{7} t+C
\end{aligned}
