Question

If the integral $\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}$, then find the value of $k$.
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

In this question, in order to the value of $k$ given that integral $\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}$, we have to first simplify the integrand by substituting $x-1=\sqrt{3}\tan y$, then in the give integral the lower limit and upper limit of the variable $x$ should be changed $y$ by putting the value $x=1$ and $x=2$ in $x-1=\sqrt{3}\tan y$ to find the respective lower limit and upper limit of the variable when we are changing the variable from $x$ to $y$. Also we have to determine the value of $dx$ in terms of the variable $y$ and $dy$. We will then evaluate the simplified integral in terms of variable $y$.

Complete step by step answer:
Let $I$ denote the integral $\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}$.
That is, let $I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}...........(1)$.
Now first we will factorise the expression ${{x}^{2}}-2x+4$ by splitting the terms.
Then we have

& {{x}^{2}}-2x+4={{x}^{2}}-2x+1+3 \\\ & ={{\left( x-1 \right)}^{2}}+3..........(2) \end{aligned}$$ Now on substituting the value of equation (2) in equation (1), we get $$I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+3 \right)}^{\dfrac{3}{2}}}}}...........(3)$$ Now let us suppose that $$x-1=\sqrt{3}\tan y$$. Now on differentiate $$x-1=\sqrt{3}\tan y$$ where differentiation of $$\tan y$$ is equals to $${{\sec }^{2}}y$$ in order to determine the value of $$dx$$ in terms of the variable $$y$$ and $$dy$$, we will get $$dx=\sqrt{3}{{\sec }^{2}}ydy......(4)$$. Now we will evaluate the lower limit of the integral $$I$$ by the value $$x=1$$ in $$x-1=\sqrt{3}\tan y$$. Putting the value the value $$x=1$$ in $$x-1=\sqrt{3}\tan y$$ , we get $$\begin{aligned} & 1-1=\sqrt{3}\tan y \\\ & \Rightarrow 0=\sqrt{3}\tan y \\\ & \Rightarrow 0=\tan y \\\ \end{aligned}$$ Since we have $$\tan y=0$$ when $$y=0$$, thus the lower limit of the variable $$y$$ is $$0$$. We will now calculate the upper limit of the integral $$I$$ by the value $$x=2$$ in $$x-1=\sqrt{3}\tan y$$. Putting the value the value $$x=2$$ in $$x-1=\sqrt{3}\tan y$$ , we get $$\begin{aligned} & 2-1=\sqrt{3}\tan y \\\ & \Rightarrow 1=\sqrt{3}\tan y \\\ & \Rightarrow \tan y=\dfrac{1}{\sqrt{3}} \\\ \end{aligned}$$ Since we have $$\tan y=\dfrac{1}{\sqrt{3}}$$ when $$y=\dfrac{\pi }{6}$$, thus the lower limit of the variable $$y$$ is $$\dfrac{\pi }{6}$$. Now on equation (4) in equation (3) and by changing the lower limit and upper limit of variable $$y$$ in equation (3), we will get $$\begin{aligned} & I=\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+3 \right)}^{\dfrac{3}{2}}}}} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left( 3{{\tan }^{2}}y+3 \right)}^{\dfrac{3}{2}}}}dy} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3\left( {{\tan }^{2}}y+1 \right) \right]}^{\dfrac{3}{2}}}}dy} \end{aligned}$$ Using $$1+{{\tan }^{2}}y={{\sec }^{2}}y$$ in the above integral, we get $$\begin{aligned} & I==\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3\left( {{\tan }^{2}}y+1 \right) \right]}^{\dfrac{3}{2}}}}dy} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{{{\left[ 3{{\sec }^{2}}y \right]}^{\dfrac{3}{2}}}}dy} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{\sqrt{3}{{\sec }^{2}}y}{3\sqrt{3}{{\sec }^{3}}y}dy} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3\sec y}dy} \end{aligned}$$ We will now use the identity $$\cos y=\dfrac{1}{\sec y}$$ in the above integral. That is on substituting $$\cos y=\dfrac{1}{\sec y}$$ in the above integral, we will have $$\begin{aligned} & I=\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3\sec y}dy} \\\ & =\int\limits_{0}^{\dfrac{\pi }{6}}{\dfrac{1}{3}\cos ydy} \\\ & =\dfrac{1}{3}\int\limits_{0}^{\dfrac{\pi }{6}}{\cos ydy} \\\ & =\dfrac{1}{3}\left[ \sin y \right]_{0}^{\dfrac{\pi }{6}} \\\ & =\dfrac{1}{3}\left[ \sin \dfrac{\pi }{6}-\sin 0 \right] \\\ & =\dfrac{1}{3}\left[ \dfrac{1}{2}-0 \right] \\\ & =\dfrac{1}{6} \end{aligned}$$ Therefore we have $$\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{1}{6}$$. But since it is given that $$\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}=\dfrac{k}{k+5}$$, therefore we have $$\dfrac{k}{k+5}=\dfrac{1}{6}$$ By cross multiplication of the above equation, we get $$\begin{aligned} & 6k=k+5 \\\ & \Rightarrow 6k-k=5 \\\ & \Rightarrow 5k=5 \\\ & \Rightarrow k=1 \end{aligned}$$ Therefore we have that the value of $$k$$ is equal to 1. **So, the correct answer is “Option A”.** **Note:** In this problem, while evaluating the integrals $$\int\limits_{1}^{2}{\dfrac{dx}{{{\left( {{x}^{2}}-2x+4 \right)}^{\dfrac{3}{2}}}}}$$ we are changing the integral in terms of variable $$y$$ using $$x-1=\sqrt{3}\tan y$$. Please take care of the fact that while evaluating the integral in terms of variable $$y$$ we have to change everything from variable $$x$$ to $$y$$ including the lower limit and the upper limit of the integral.