Question

What is the integral of ${{\cos }^{5}}x$ ?

Answer 1

In this problem, we need to find the integral of ${{\cos }^{5}}x$ . For this, we write it as ${{\cos }^{4}}x.\cos x$ and then as ${{\left( {{\cos }^{2}}x \right)}^{2}}.\cos x$ . After that, using the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , we can rewrite it as $\int{{{\left( 1-{{\sin }^{2}}x \right)}^{2}}.\cos xdx}$ . We now assume $\sin x=t$ and rewrite it as $\int{{{\left( 1-{{t}^{2}} \right)}^{2}}dt}$ . Finally, solving the integration and substituting t back, we get, $\sin x-\dfrac{2}{3}{{\left( \sin x \right)}^{3}}+\dfrac{{{\left( \sin x \right)}^{5}}}{5}+c$ .

Complete step-by-step answer:
In this problem, we need to find out the integral of ${{\cos }^{5}}x$ . We can write the integral and then equate it to I as,
$I=\int{{{\cos }^{5}}xdx}$
We can break down the term ${{\cos }^{5}}x$ as ${{\cos }^{4}}x.\cos x$ which can also be rewritten as ${{\left( {{\cos }^{2}}x \right)}^{2}}.\cos x$ . The integral thus becomes,
$\Rightarrow I=\int{{{\left( {{\cos }^{2}}x \right)}^{2}}.\cos xdx}$
om our knowledge of trigonometric identity, we know that
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
From this identity, we get,
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x$
We now put this value of ${{\cos }^{2}}x$ in the integral. The integral thus becomes,
$\Rightarrow I=\int{{{\left( 1-{{\sin }^{2}}x \right)}^{2}}.\cos xdx}....\left( i \right)$
Let us now assume $\sin x=t$ $\Rightarrow \cos xdx=dt$ . Substituting these in integral (i), we get,
$\Rightarrow I=\int{{{\left( 1-{{t}^{2}} \right)}^{2}}dt}$
We know apply the formula of squares ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ in the above integral and get,
$\Rightarrow I=\int{\left( 1-2{{t}^{2}}+{{t}^{4}} \right)dt}$
We now break down the integral into its sub-parts as,
$\Rightarrow I=\int{dt}-\int{2{{t}^{2}}dt}+\int{{{t}^{4}}dt}$
Performing the integration using the standard formulae according to our knowledge, we get,
$\Rightarrow I=t-\dfrac{2}{3}{{t}^{3}}+\dfrac{{{t}^{5}}}{5}+c$ where c is the integration constant.
Now, substituting t back in the above equation, we get,
$\Rightarrow I=\sin x-\dfrac{2}{3}{{\left( \sin x \right)}^{3}}+\dfrac{{{\left( \sin x \right)}^{5}}}{5}+c$
Therefore, we can conclude that the integral of ${{\cos }^{5}}x$ is $\sin x-\dfrac{2}{3}{{\left( \sin x \right)}^{3}}+\dfrac{{{\left( \sin x \right)}^{5}}}{5}+c$ .

Note: We should be careful over here as this process involves integration by substitution. A lot of substitutions are involved in this problem, so we need to be extra cautious and carry out the substitutions. In the end, we should remember to substitute back the variables to the original variable.