Question

Evaluate the following integral \[\int{\left( x-5 \right)\sqrt{{{x}^{2}}-1}dx}\]

Answer 1

Hint: Here, we have to evaluate \[\int{\left( x-5 \right)\sqrt{{{x}^{2}}-1}dx}\]. For this, we will first assume it to be ‘I’ and then expand the inside of the integral. Then, we will separate both the functions formed and hence form two separate integrals. Then we will assume those separate integrals to be $ {{I}_{1}} $ and $ {{I}_{2}} $ . Then, we will solve both of the integrals separately by using the basic properties and identities of integration. Once we get the values of both $ {{I}_{1}} $ and $ {{I}_{2}} $ , we will get the value of I. Hence, we will have an answer.

Complete step by step answer:
Now, here we have to evaluate the integral \[\int{\left( x-5 \right)\sqrt{{{x}^{2}}-1}dx}\].
For this, let us first assume this integral to be ‘I’. Thus, we can say that:
 $ I=\int{\left( x-5 \right)\sqrt{{{x}^{2}}-1}dx} $
Now, we will first solve the inside of the integral. This will give us:
 $ \begin{align}
  & I=\int{\left( x-5 \right)\sqrt{{{x}^{2}}-1}dx} \\
 & \Rightarrow I=\int{\left( x\sqrt{{{x}^{2}}-1}-5\sqrt{{{x}^{2}}-1} \right)dx} \\
\end{align} $
Now, we know that $ \int{\left( f\left( x \right)-g\left( x \right) \right)dx=}\int{f\left( x \right)dx-}\int{g\left( x \right)dx} $
Thus, we can write I as:
 $ \begin{align}
  & I=\int{\left( x\sqrt{{{x}^{2}}-1}-5\sqrt{{{x}^{2}}-1} \right)dx} \\
 & \Rightarrow I=\int{x\sqrt{{{x}^{2}}-1}dx}-\int{5\sqrt{{{x}^{2}}-1}dx} \\
\end{align} $
Now, let us assume that $ \int{x\sqrt{{{x}^{2}}-1}dx} $ is equal to $ {{I}_{1}} $ and $ \int{5\sqrt{{{x}^{2}}-1}dx} $ is equal to $ {{I}_{2}} $ .
Thus, we can write I as:
 $ I={{I}_{1}}-{{I}_{2}} $ …..(i)
Now, we will first solve $ {{I}_{1}} $ .
We have:
 $ {{I}_{1}}=\int{x\sqrt{{{x}^{2}}-1}dx} $
Now, let us assume that $ \sqrt{{{x}^{2}}-1} $ is equal to ‘t’.
Thus, we have:
 $ \sqrt{{{x}^{2}}-1}=t $
Now, squaring both sides, we get:
 $ \begin{align}
  & \sqrt{{{x}^{2}}-1}=t \\
 & \Rightarrow {{x}^{2}}-1={{t}^{2}} \\
\end{align} $
Now, differentiating both sides, we will get:
 $ \begin{align}
  & {{x}^{2}}-1={{t}^{2}} \\
 & \Rightarrow 2x.dx=2t.dt \\
 & \Rightarrow xdx=tdt \\
\end{align} $
Now, putting the value of ‘xdx’ and $ \sqrt{{{x}^{2}}-1} $ in $ {{I}_{1}} $ we get:
 $ \begin{align}
  & {{I}_{1}}=\int{x\sqrt{{{x}^{2}}-1}dx} \\
 & \Rightarrow {{I}_{1}}=\int{t.tdt} \\
 & \Rightarrow {{I}_{1}}=\int{{{t}^{2}}dt} \\
\end{align} $
Now, we know that $ \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1} $ . Thus, we have the value of $ {{I}_{1}} $ as:
 $ \begin{align}
  & {{I}_{1}}=\int{{{t}^{2}}dt} \\
 & \Rightarrow {{I}_{1}}=\dfrac{{{t}^{2+1}}}{2+1} \\
 & \Rightarrow {{I}_{1}}=\dfrac{{{t}^{3}}}{3} \\
\end{align} $
Now, putting the value of ‘t’ in the final value of $ {{I}_{1}} $ we get:
 $ \begin{align}
  & {{I}_{1}}=\dfrac{{{t}^{3}}}{3} \\
 & \Rightarrow {{I}_{1}}=\dfrac{{{\left( \sqrt{{{x}^{2}}-1} \right)}^{3}}}{3} \\
 & \Rightarrow {{I}_{1}}=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}{3} \\
\end{align} $
Now, we will find the value of $ {{I}_{2}} $ .
We have:
 $ {{I}_{2}}=\int{5\sqrt{{{x}^{2}}-1}dx} $
Now, we know that $ \int{kf\left( x \right)dx}=k\int{f\left( x \right)dx} $ where ‘k’ is a constant.
Thus, we have $ {{I}_{2}} $ as:
 $ \begin{align}
  & {{I}_{2}}=\int{5\sqrt{{{x}^{2}}-1}dx} \\
 & \Rightarrow {{I}_{2}}=5\int{\sqrt{{{x}^{2}}-1}dx} \\
\end{align} $
Now, we know that $ \int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right) $
Here, we have a=1.
Thus, we get the value of $ {{I}_{2}} $ as:
 $ \begin{align}
  & {{I}_{2}}=5\int{\sqrt{{{x}^{2}}-1}dx} \\
 & \Rightarrow {{I}_{2}}=5\left( \dfrac{x}{2}\sqrt{{{x}^{2}}-1}-\dfrac{1}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right) \right) \\
 & \Rightarrow {{I}_{2}}=\dfrac{5x}{2}\sqrt{{{x}^{2}}-1}-\dfrac{5}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right) \\
\end{align} $
Now, we have the value of both $ {{I}_{1}} $ and $ {{I}_{2}} $ . Thus, putting their values in equation (i) we get:
 $ \begin{align}
  & I={{I}_{1}}-{{I}_{2}} \\
 & \Rightarrow I=\left( \dfrac{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}{3} \right)-\left( \dfrac{5x}{2}\sqrt{{{x}^{2}}-1}-\dfrac{5}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right) \right) \\
 & \Rightarrow I=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}{3}-\dfrac{5x}{2}\sqrt{{{x}^{2}}-1}+\dfrac{5}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right) \\
\end{align} $
Now, since it is indefinite integration, we need to add a constant of integration ‘C’ along with it. Thus, we get the final value of I as:
 $ I=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}{3}-\dfrac{5x}{2}\sqrt{{{x}^{2}}-1}+\dfrac{5}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-1} \right)+C $
Thu, we have our answer.

Note:
Different identities of integration are given as follows which may come in handy:
1. $ \int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right) $
2. $ \int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right) $
3. $ \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}-\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a} $
Make sure to start solving by opening the bracket and considering the function as a sum of two functions rather than taking it as a product of two functions and trying to apply the integration by parts. It will take time and the solution will be complex.