Question

Evaluate $\int{{{\sin }^{7}}xdx}$.

Answer 1

Hint: Clearly, the evaluation is tough with such high degree so we will split 7 into 6 and 1 pair after that we will use the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. After getting to this point we will use ${{\left( a-b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ formula and solve it properly. Finally we will take help of substitution $p=\cos x$ to solve it further.

Complete step by step solution:
We are given the term $\int{{{\sin }^{7}}xdx}$ and we need to find its simplest evaluation. So, for this we are going to split 7 into 1 and 6 in the trigonometric term, ${{\sin }^{7}}x={{\sin }^{6}}x\cdot {{\sin }^{1}}x$. This will result into a new trigonometric form $\int{{{\sin }^{7}}}xdx=\int{{{\sin }^{6}}x}\cdot {{\sin }^{1}}xdx=\int{{{\left( \sin x \right)}^{6}}\cdot \sin x}dx$…(i).
We can now convert the term ${{\left( \sin x \right)}^{6}}={{\left( \sin x \right)}^{2\times 3}}={{\left( {{\sin }^{2}}x \right)}^{3}}$ and substitute it in (i). Therefore, we get $\int{{{\sin }^{7}}xdx=\int{{{\left( {{\sin }^{2}}x \right)}^{3}}\cdot \sin x}dx}$…(ii).
As we know the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ we can take out the value of sin as ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ and put it in place of ${{\sin }^{2}}x$. Thus, we get $\int{{{\sin }^{7}}xdx=\int{{{\left( 1-{{\cos }^{2}}x \right)}^{3}}\cdot \sin x}dx}$.
After this we will use algebra in which we will take the help of formula ${{\left( a-b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$. By considering $a=1,b=-{{\cos }^{2}}x$ we get
\[\begin{align}
  & \int{{{\sin }^{7}}xdx=\int{{{\left( 1-{{\cos }^{2}}x \right)}^{3}}\cdot \sin x}dx} \\
 & \Rightarrow \int{{{\sin }^{7}}xdx=\int{\left[ {{\left( 1 \right)}^{3}}+{{\left( -{{\cos }^{2}}x \right)}^{3}}+3{{\left( 1 \right)}^{2}}\left( -{{\cos }^{2}}x \right)+3\left( 1 \right){{\left( -{{\cos }^{2}}x \right)}^{2}} \right]\cdot \sin x}dx} \\
 & \Rightarrow \int{{{\sin }^{7}}xdx=\int{\left[ \sin x+{{\left( -{{\cos }^{2}}x \right)}^{3}}\cdot \sin x+3\left( -{{\cos }^{2}}x \right)\cdot \sin x+3{{\left( -{{\cos }^{2}}x \right)}^{2}}\cdot \sin x \right]}dx} \\
 & \Rightarrow \int{{{\sin }^{7}}xdx=\int{\left[ \sin x-{{\left( \cos x \right)}^{6}}\cdot \sin x-3{{\left( \cos x \right)}^{2}}\cdot \sin x+3{{\left( \cos x \right)}^{4}}\cdot \sin x \right]}dx} \\
 & \Rightarrow \int{{{\sin }^{7}}xdx=\int{\sin x}}dx-\int{{{\left( \cos x \right)}^{6}}\cdot \sin x}dx-3\int{{{\left( \cos x \right)}^{2}}\cdot \sin x}dx+3\int{{{\left( \cos x \right)}^{4}}\cdot \sin x}dx\,\,\,...(iii) \\
\end{align}\]
Now, it is time to use the process of substitution here. In this process we basically put one term in place of the other. And here, we will put $p=\cos x$. After applying differentiation with respect to x, we get,
$\begin{align}
  & p=\cos x \\
 & \Rightarrow \dfrac{d}{dx}\left( p \right)=\dfrac{d}{dx}\left( \cos x \right) \\
 & \Rightarrow \dfrac{dp}{dx}=-\sin x\,\,\left[ \text{Since, }\dfrac{d\left( \cos x \right)}{dx}=-\sin x \right] \\
 & \Rightarrow dp=-\sin xdx \\
\end{align}$
Substituting all above differentiation in (iii), we have
\[\int{{{\sin }^{7}}xdx=-\int{dp}}+\int{{{p}^{6}}}dp+3\int{{{p}^{2}}}dp-3\int{{{p}^{4}}}dp\]
By using the integration $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ results into $\int{dp}=p,\int{{{p}^{6}}dp}=\dfrac{{{p}^{7}}}{7},\int{{{p}^{2}}dp}=\dfrac{{{p}^{3}}}{3},\int{{{p}^{4}}dp=\dfrac{{{p}^{5}}}{5}}$. Therefore,
\[\begin{align}
  & \Rightarrow \int{{{\sin }^{7}}xdx}=p+\dfrac{{{p}^{7}}}{7}+3\times \dfrac{{{p}^{3}}}{3}-3\times \dfrac{{{p}^{5}}}{5} \\
 & \Rightarrow \int{{{\sin }^{7}}xdx}=p+\dfrac{{{p}^{7}}}{7}+{{p}^{3}}-\dfrac{3{{p}^{5}}}{5} \\
\end{align}\]
Since, $p=\cos x$ thus, \[\int{{{\sin }^{7}}xdx=-\cos x}+\dfrac{{{\cos }^{7}}x}{7}+{{\cos }^{3}}x-\dfrac{3{{\cos }^{5}}x}{5}\].

Note: To solve such types of questions we will split the odd degree into even degrees. Here, after splitting the odd degree term we must get one even degree and the other as degree = 1. We have converted ${{\left( \sin x \right)}^{6}}={{\left( \sin x \right)}^{2\times 3}}={{\left( {{\sin }^{2}}x \right)}^{3}}$ and not ${{\left( \sin x \right)}^{6}}={{\left( \sin x \right)}^{2\times 3}}={{\left( {{\sin }^{3}}x \right)}^{2}}$ there is the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ which is off degree 2 and not 3. During the process of substitution we could have used $p=-\cos x$ instead of $p=\cos x$. In $p=-\cos x$ we get,
$\begin{align}
  & p=-\cos x \\
 & \Rightarrow \dfrac{d}{dx}\left( p \right)=-\dfrac{d}{dx}\left( \cos x \right) \\
 & \Rightarrow \dfrac{dp}{dx}=-\left( -\sin x \right)\,\,\left[ \text{Since, }\dfrac{d\left( \cos x \right)}{dx}=-\sin x \right] \\
 & \Rightarrow dp=\sin xdx \\
\end{align}$
These after differentiation may give slightly different terms but lastly will lead to the same answer which is right. But if the focus gets deviated then we might get the wrong answer, so it is important to solve it with concentration.