Question

How to Integrate $\dfrac{7}{2}$ & $\dfrac{1}{2}$ of $\dfrac{1}{{{\left( 5+4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx$?

Answer 1

This question is from the topic of integration. In solving this question, we will use a substitution method to solve this question. First, we make the term $\left( 5+4x-{{x}^{2}} \right)$ simple. After that, we will take $\left( x-2 \right)$ as $3\sin t$ and differentiate them. After that, we use them for further integration. After solving the further integration, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
We can write $5+4x-{{x}^{2}}$ as
$5+4x-{{x}^{2}}=9-\left( 4-4x+{{x}^{2}} \right)$
Using the formula ${{\left( a-b \right)}^{2}}={{\left( b-a \right)}^{2}}={{a}^{2}}-2\times a\times b+{{b}^{2}}$, we can write the above as
$\Rightarrow 5+4x-{{x}^{2}}=9-{{\left( x-2 \right)}^{2}}$
The above can also be written as
$\Rightarrow 5+4x-{{x}^{2}}={{3}^{2}}-{{\left( x-2 \right)}^{2}}$
Now, we will do the substitution.
Let us take $\left( x-2 \right)$ as $3\sin t$
So, we can write
$\left( x-2 \right)=3\sin t$
Now, differentiating both sides of the above equation, we get
$dx=3\cos tdt$
Now, we know that we have to integrate the term $\dfrac{1}{{{\left( 5+4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx$ having limits as $\dfrac{7}{2}$ and $\dfrac{1}{2}$. So, let us write
$I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{1}{{{\left( 5+4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx}$
The above can also be written as
$\Rightarrow I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{1}{{{\left( {{3}^{2}}-{{\left( x-2 \right)}^{2}} \right)}^{\dfrac{3}{2}}}}dx}$
Now, putting the values of $\left( x-2 \right)$ as $3\sin t$ and $dx$ as $3\cos tdt$ that we have found above, we can write
$\Rightarrow I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{1}{{{\left( {{3}^{2}}-{{\left( 3\sin t \right)}^{2}} \right)}^{\dfrac{3}{2}}}}3\cos tdt}$
We are changing the variable, so limits will also change according to them. We can write
At $x=\dfrac{7}{2}$,
$\left( \dfrac{7}{2}-2 \right)=3\sin t$
$\Rightarrow \left( \dfrac{1}{2} \right)=\sin t$
$\Rightarrow t=\dfrac{\pi }{6}$
So, at $x=\dfrac{7}{2}$, $t=\dfrac{\pi }{6}$
And at $x=\dfrac{1}{2}$,
$\left( \dfrac{1}{2}-2 \right)=3\sin t$
$\Rightarrow \left( -\dfrac{3}{2} \right)=3\sin t$
$\Rightarrow \left( -\dfrac{1}{2} \right)=\sin t$
$\Rightarrow t=-\dfrac{\pi }{6}$
So, at $x=\dfrac{1}{2}$, $t=-\dfrac{\pi }{6}$
Now, we can write $I=\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{1}{{{\left( {{3}^{2}}-{{\left( 3\sin t \right)}^{2}} \right)}^{\dfrac{3}{2}}}}3\cos tdt}$ as
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{\left( {{3}^{2}}-{{\left( 3\sin t \right)}^{2}} \right)}^{\dfrac{3}{2}}}}3\cos tdt}$
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{\left( {{3}^{2}}-{{3}^{2}}{{\sin }^{2}}t \right)}^{\dfrac{3}{2}}}}3\cos tdt}$
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2\times \dfrac{3}{2}}}{{\left( 1-{{\sin }^{2}}t \right)}^{\dfrac{3}{2}}}}3\cos tdt}$
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{3}}{{\left( 1-{{\sin }^{2}}t \right)}^{\dfrac{3}{2}}}}3\cos tdt}$
The above can also be written as
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2}}{{\left( 1-{{\sin }^{2}}t \right)}^{\dfrac{3}{2}}}}\cos tdt}$
Using the formula $\left( 1-{{\sin }^{2}}t \right)={{\cos }^{2}}t$, we can write
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2}}{{\left( {{\cos }^{2}}t \right)}^{\dfrac{3}{2}}}}\cos tdt}$
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2}}{{\cos }^{2\times \dfrac{3}{2}}}t}\cos tdt}$
$\Rightarrow I=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2}}{{\cos }^{3}}t}\cos tdt}$
The above can also be written as
$\Rightarrow I=\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{3}^{2}}{{\cos }^{2}}t}dt}$
$\Rightarrow I=\dfrac{1}{{{3}^{2}}}\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{\cos }^{2}}t}dt}$
$\Rightarrow I=\dfrac{1}{9}\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{1}{{{\cos }^{2}}t}dt}$
$\sec x=\dfrac{1}{\cos x}$
Using the formula $\sec x=\dfrac{1}{\cos x}$, we can the above as
$\Rightarrow I=\dfrac{1}{9}\int\limits_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{{{\sec }^{2}}tdt}$
Using the formula $\int{{{\sec }^{2}}xdx}=\tan x$, we can write the above as
$\Rightarrow I=\dfrac{1}{9}\left[ \tan t \right]_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}$
The above can be written as
$\Rightarrow I=\dfrac{1}{9}\left[ \tan \dfrac{\pi }{6}-\tan \left( -\dfrac{\pi }{6} \right) \right]$
Now, using the formula $\tan \left( -x \right)=-\tan x$, we can write
$\Rightarrow I=\dfrac{1}{9}\left[ \tan \dfrac{\pi }{6}+\tan \dfrac{\pi }{6} \right]$
The above can be written as
$\Rightarrow I=\dfrac{2}{9}\left[ \tan \dfrac{\pi }{6} \right]$
Using the formula $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$, we can write
$\Rightarrow I=\dfrac{2}{9}\dfrac{1}{\sqrt{3}}=\dfrac{2}{9\sqrt{3}}$
Hence, we have integrated the term $\int\limits_{\dfrac{1}{2}}^{\dfrac{7}{2}}{\dfrac{1}{{{\left( 5+4x-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}dx}$ and we have got our answer as $\dfrac{2}{9\sqrt{3}}$

Note: We should have a better knowledge in the topic of integration for solving this type of question.
Remember the following formulas to solve this type of question easily:
$\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$
$\tan \left( -x \right)=-\tan x$
$\int{{{\sec }^{2}}xdx}=\tan x$
$\sec x=\dfrac{1}{\cos x}$
${{\left( a-b \right)}^{2}}={{\left( b-a \right)}^{2}}={{a}^{2}}-2\times a\times b+{{b}^{2}}$
$\left( 1-{{\sin }^{2}}t \right)={{\cos }^{2}}t$