Question

How do you evaluate the given indefinite integral $\int{{{\sin }^{3}}x{{\cos }^{5}}xdx}$?

Answer 1

We start solving the problem by making use of the fact that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the given integrand. We then make the necessary calculations and then assume $y=\cos x$ to proceed through the problem. We then find $dy$ in terms of $dx$ by making the necessary calculations. We then make use of the facts that $\int{\left( a-b \right)dx}=\int{adx}+\int{bdx}$, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$ to proceed through the problem. We then substitute $y=\cos x$ to get the required answer for the given indefinite integral.

Complete step-by-step answer:
According to the problem, we are asked to evaluate the given indefinite integral $\int{{{\sin }^{3}}x{{\cos }^{5}}xdx}$.
Let us assume $I=\int{{{\sin }^{3}}x{{\cos }^{5}}xdx}$.
$\Rightarrow I=\int{\sin x\times \left( {{\sin }^{2}}x \right)\times {{\cos }^{5}}xdx}$ ---(1).
We know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$. Let us use this result in equation (1).
$\Rightarrow I=\int{\sin x\times \left( 1-{{\cos }^{2}}x \right)\times {{\cos }^{5}}xdx}$.
$\Rightarrow I=\int{\left( {{\cos }^{5}}x-{{\cos }^{7}}x \right)\times \sin xdx}$.
$\Rightarrow I=\int{\left( {{\cos }^{7}}x-{{\cos }^{5}}x \right)\times \left( -\sin x \right)dx}$ ---(2).
Let us assume $y=\cos x$ ---(3).
Let us apply a differential on both sides of equation (3).
$\Rightarrow d\left( y \right)=d\left( \cos x \right)$.
$\Rightarrow dy=-\sin xdx$ ---(4).
Let us substitute equations (3) and (4) in equation (2).
$\Rightarrow I=\int{\left( {{y}^{7}}-{{y}^{5}} \right)dy}$ ---(5).
We know that $\int{\left( a-b \right)dx}=\int{adx}+\int{bdx}$. Let us use this result in equation (5).
$\Rightarrow I=\int{{{y}^{7}}dy}-\int{{{y}^{5}}dy}$ ---(6).
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$. Let us use this result in equation (6).
$\Rightarrow I=\dfrac{{{y}^{7+1}}}{7+1}-\dfrac{{{y}^{5+1}}}{5+1}+C$.
$\Rightarrow I=\dfrac{{{y}^{8}}}{8}-\dfrac{{{y}^{6}}}{6}+C$ ---(7).
Let us substitute equation (3) in equation (7).
$\Rightarrow I=\dfrac{{{\cos }^{8}}x}{8}-\dfrac{{{\cos }^{6}}x}{6}+C$.
So, we have found the result of integration of indefinite integral $\int{{{\sin }^{3}}x{{\cos }^{5}}xdx}$ as $\dfrac{{{\cos }^{8}}x}{8}-\dfrac{{{\cos }^{6}}x}{6}+C$.
$\therefore$ The result of integration of indefinite integral $\int{{{\sin }^{3}}x{{\cos }^{5}}xdx}$ as $\dfrac{{{\cos }^{8}}x}{8}-\dfrac{{{\cos }^{6}}x}{6}+C$.

Note: We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully to avoid confusion and calculation mistakes. We should not forget to add constant integration while solving this type of problem. We can also solve this problem by making use of the reduction formula of $\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}$. Similarly, we can expect problems to find the value of definite integral $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}x{{\cos }^{5}}xdx}$.