Answer
$$\text{a) }\displaystyle{\int\cos^nx\ dx=\frac{1}{n}\times\cos^{n-1}x\sin x+\frac{n-1}{n}\int cos^{n-2}x\ dx}\\
\text{b) }\displaystyle{\int\cos^2x\ dx=\frac{1}{4}\sin {2x}+\frac{1}{2}x+C}\\
\text{c) }\displaystyle{\int\cos^4x\ dx=\frac{1}{8}\sin2x\cos^2x+\frac{3}{16}\sin {2x}+\frac{3}{8}x+C}$$
Work Step by Step
a)
$I=\displaystyle \int\cos^nx\ dx$
$\displaystyle \left[\begin{array}{ll} u=\cos^{n-1}x & dv=\cos x \\ & \\ du=-(n-1)\cos^{n-2}x\sin x & v=\sin x \end{array}\right]$ Integration by parts
$I=\displaystyle{\cos^{n-1}x\sin x-\int-(n-1)cos^{n-2}x\sin x\sin x\ dx}\\
I=\displaystyle{\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x\sin^2x\ dx}$
$$\sin^2x=1-\cos^2x$$
$\displaystyle{I=\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x(1-\cos^2x)\ dx}\\
I=\displaystyle{\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x-\cos^nx\ dx}\\
\displaystyle{I=\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x\ dx-(n-1)\int\cos^nx\ dx}\\
\displaystyle{\int\cos^nx\ dx=\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x\ dx-(n-1)\int\cos^nx\ dx}\\
\displaystyle{\int\cos^nx\ dx-(n-1)+\int\cos^nx\ dx=\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x\ dx}\\
\displaystyle{n\int\cos^nx\ dx=\cos^{n-1}x\sin x+(n-1)\int cos^{n-2}x\ dx}\\
\displaystyle{\int\cos^nx\ dx=\frac{1}{n}\times\cos^{n-1}x\sin x+\frac{n-1}{n}\int cos^{n-2}x\ dx}\\$
b)
$\displaystyle{\int\cos^2x\ dx=\frac{1}{2}\cos x\sin x+\frac{1}{2}\int cos^{0}x\ dx}\\
\displaystyle{\int\cos^2x\ dx=\frac{1}{2}\times\frac{1}{2}\times2\cos x\sin x+\frac{1}{2}\int1\ dx}\\
\displaystyle{\int\cos^2x\ dx=\frac{1}{4}\sin {2x}+\frac{1}{2}x+C}$
c)
$\displaystyle{\int\cos^4x\ dx=\frac{1}{4}\cos^3x\sin x+\frac{3}{4}\int\cos^{2}x\ dx}\\
\displaystyle{\int\cos^4x\ dx=\frac{1}{4}\times\frac{1}{2}\times2\cos x\sin x\cos^2x+\frac{3}{4}\left(\frac{1}{4}\sin {2x}+\frac{1}{2}x\right)+C}\\
\displaystyle{\int\cos^4x\ dx=\frac{1}{8}\sin2x\cos^2x+\frac{3}{16}\sin {2x}+\frac{3}{8}x+C}$