Question

a) Prove the reduction formula \(\)\(\int {{{\cos }^n}xdx = \frac{1}{n}{{\cos }^{n - 1}}x\sin x + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}xdx} } \)

b)Use part a) to evaluate \(\int {{{\cos }^2}xdx} \)

c) Use parts a) and b) to evaluate \(\int {{{\cos }^4}xdx} \)

Short answer

We should use the substitution method to prove the integral

Let’s take \(\begin{aligned}{l}\mu = {\cos ^{n - 1}}x\\dv = \cos xdx\end{aligned}\)

Step-by-step solution

Given Data

To prove: \(\int {{{\cos }^n}xdx = \frac{1}{n}{{\cos }^{n - 1}}x\sin x + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}xdx} } \)

Let’s take \(\begin{aligned}{l}\mu = {\cos ^{n - 1}}x\\d\mu = \left( {n - 1} \right){\cos ^{n - 1}}x\sin xdx\end{aligned}\) \(\begin{aligned}{l}dv = \cos xdx\\v = \sin x\end{aligned}\)

\(\begin{aligned}{l}\int {{{\cos }^n}xdx = {{\cos }^{n - 1}}x\sin x + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x{{\sin }^2}xdx} } \\ = {\cos ^{n - 1}}x\sin x + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right)dx} \\ = {\cos ^{n - 1}}x\sin x + \left( {n - 1} \right)\int {{{\cos }^{n - 2}}xdx - \left( {n - 1} \right){{\int {\cos } }^n}xdx} \end{aligned}\)

Rearranging Terms

\(\begin{aligned}{l}n\int {{{\cos }^n}xdx = {{\cos }^{n - 1}}x\sin x + \left( {n - 1} \right)\int {{{\cos }^{\left( {n - 2} \right)}}xdx} } \\\int {{{\cos }^n}xdx = \frac{1}{n}} {\cos ^{n - 1}}x\sin x + \frac{{n - 1}}{n}\int {{{\cos }^{\left( {n - 2} \right)}}xdx} \end{aligned}\)

LHS=RHS

Hence proved.

Answer: (b)

Calculation

Take\(n = 2\)in part a), to get

\(\begin{aligned}{l}\int {{{\cos }^2}xdx} = \frac{1}{2}\cos x\sin x + \frac{1}{2}\int {1dx} \\\therefore \int {{{\cos }^2}xdx} = \frac{x}{2} + \frac{{\sin 2x}}{4} + c\end{aligned}\)

Rearranging Terms

\(\begin{aligned}{l}\int {{{\cos }^4}xdx} = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{4}\int {{{\cos }^2}xdx} \\\therefore \int {{{\cos }^4}xdx} = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{8}x + \frac{3}{{16}}\sin 2x + c\end{aligned}\)

Hence, from parts a) and b)

\(\int {{{\cos }^4}xdx} = \frac{1}{4}{\cos ^3}x\sin x + \frac{3}{8}x + \frac{3}{{16}}\sin 2x + c\)