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

Hint: 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:
\[\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}\]
\[\tan \left( -x \right)=-\tan x\]
\[\int{{{\sec }^{2}}xdx}=\tan x\]
\[\sec x=\dfrac{1}{\cos x}\]
\[{{\left( a-b \right)}^{2}}={{\left( b-a \right)}^{2}}={{a}^{2}}-2\times a\times b+{{b}^{2}}\]
\[\left( 1-{{\sin }^{2}}t \right)={{\cos }^{2}}t\]