Question

How do you integrate $\int{{{\sin }^{5}}\left( x \right){{\cos }^{2}}\left( x \right)dx}$?

Answer 1

In order to find the solution to the given question, that is to integrate $\int{{{\sin }^{5}}\left( x \right){{\cos }^{2}}\left( x \right)dx}$, apply one of the trigonometric identities which is ${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$. With the help of this identity, we convert the whole function in terms of cosine. After this assume $u=\cos \left( x \right)$, and replace all the cosine terms with $u$. Then simplify the integral with the help of integral formula that is $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$.

Complete step by step solution:
According to the question, given function which needs to be integrated is as follows:
$\int{{{\sin }^{5}}\left( x \right){{\cos }^{2}}\left( x \right)dx}$
We can rewrite the above integral in the following equivalent statements:
$\Rightarrow \int{\sin \left( x \right){{\sin }^{4}}\left( x \right){{\cos }^{2}}\left( x \right)dx}$
The above statements can again be rewritten as:
$\Rightarrow \int{{{\left( {{\sin }^{2}}\left( x \right) \right)}^{2}}{{\cos }^{2}}\left( x \right)\sin \left( x \right)dx}$
Now apply one of the trigonometric identities which is ${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$ and convert the above expression in terms of cosine, we get:
$\Rightarrow \int{{{\left( 1-{{\cos }^{2}}\left( x \right) \right)}^{2}}{{\cos }^{2}}\left( x \right)\sin \left( x \right)dx}$
To make the above expression easier to integrate, let us assume $u=\cos \left( x \right)$, and replace all the cosine terms with $u$in the above expression. Also take the derivative of $u$ to replace $dx$ with $du$, we get:

& u=\cos x \\\ & \Rightarrow \dfrac{du}{dx}=-\sin \left( x \right) \\\ & \Rightarrow dx=\dfrac{du}{-\sin \left( x \right)} \\\ \end{aligned}$$ And hence we can write: $$\Rightarrow \int{\dfrac{{{\left( 1-{{u}^{2}} \right)}^{2}}{{u}^{2}}\sin \left( x \right)}{-\sin \left( x \right)}}du$$ As we can see that $$\sin \left( x \right)$$ is there in both numerator and denominator, therefore it gets cancelled and we get: $$\Rightarrow -\int{{{\left( 1-{{u}^{2}} \right)}^{2}}{{u}^{2}}}du$$ Apply the formula $${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$$, to simplify the above expression, we get: $$\Rightarrow -\int{{{u}^{2}}\left( 1+{{u}^{4}}-2{{u}^{2}} \right)}du$$ Simplify it further by opening the brackets and using multiplication in the above expression, we get: $$\Rightarrow -\int{{{u}^{2}}+{{u}^{6}}-2{{u}^{4}}}du$$ $$\Rightarrow -\int{{{u}^{2}}du-\int{{{u}^{6}}}du-\int{-2{{u}^{4}}}}du$$ To find the integration of the terms in the above expression, apply the integral formula that is $$\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$$, we get: $$\Rightarrow -\dfrac{1}{3}{{u}^{3}}-\dfrac{1}{7}{{u}^{7}}+\dfrac{2}{5}{{u}^{5}}+C$$ Now, substitute this $$u=\cos \left( x \right)$$ in the above expression where C is the integration constant, we get: $$\Rightarrow -\dfrac{1}{3}{{\cos }^{3}}\left( x \right)-\dfrac{1}{7}{{\cos }^{7}}\left( x \right)+\dfrac{2}{5}{{\cos }^{5}}\left( x \right)+C$$ $$\Rightarrow \int{{{\sin }^{5}}\left( x \right){{\cos }^{2}}\left( x \right)dx}=-\dfrac{1}{3}{{\cos }^{3}}\left( x \right)-\dfrac{1}{7}{{\cos }^{7}}\left( x \right)+\dfrac{2}{5}{{\cos }^{5}}\left( x \right)+C$$ Therefore, integration of $$\int{{{\sin }^{5}}\left( x \right){{\cos }^{2}}\left( x \right)dx}$$ is equal to $$-\dfrac{1}{3}{{\cos }^{3}}\left( x \right)-\dfrac{1}{7}{{\cos }^{7}}\left( x \right)+\dfrac{2}{5}{{\cos }^{5}}\left( x \right)+C$$. **Note:** Students make mistakes by not using the substitution method in this type of question, and solve the integral in the form of trigonometric function only, although this is also a way to solve this question. It’s way more complex and sometimes leads to the wrong answer because of calculation mistakes. So, it is better to use the substitution method as it makes the integration easier.