Mathematics

Evaluate $$\displaystyle\int\limits_4^9 {\dfrac{{dx}}{{\sqrt {\left( {9 - x} \right)\left( {x - 4} \right)} }}} $$


SOLUTION

Consider the given integral.

$$I=\int_{4}^{9}{\dfrac{dx}{\sqrt{\left( 9-x \right)\left( x-4 \right)}}}$$

$$ I=\int_{4}^{9}{\dfrac{dx}{\sqrt{9x-36-{{x}^{2}}+4x}}} $$

$$ I=\int_{4}^{9}{\dfrac{dx}{\sqrt{13x-36-{{x}^{2}}}}} $$

$$ I=\int_{4}^{9}{\dfrac{dx}{\sqrt{13x+\dfrac{169}{4}-\dfrac{169}{4}-36-{{x}^{2}}}}} $$

$$ I=\int_{4}^{9}{\dfrac{dx}{\sqrt{\dfrac{25}{4}-\left( -13x+\dfrac{169}{4}+{{x}^{2}} \right)}}} $$

$$ I=\int_{4}^{9}{\dfrac{dx}{\sqrt{{{\left( \dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{13}{2}-x \right)}^{2}}}}} $$

 

Let

$$t=\dfrac{13}{2}-x$$

$$dt=-dx$$

 

Therefore,

$$I=-\int_{\dfrac{5}{2}}^{-\dfrac{5}{2}}{\dfrac{dt}{\sqrt{{{\left( \dfrac{5}{2} \right)}^{2}}-{{t}^{2}}}}}$$

 

We know that

$$\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$$

 

Therefore,

$$ I=-\left[ {{\sin }^{-1}}\left( \dfrac{t}{\dfrac{5}{2}} \right) \right]_{\dfrac{5}{2}}^{-\dfrac{5}{2}} $$

$$ I=-\left[ {{\sin }^{-1}}\left( \dfrac{2\left( -\dfrac{5}{2} \right)}{5} \right)-{{\sin }^{-1}}\left( \dfrac{2\left( \dfrac{5}{2} \right)}{5} \right) \right] $$

$$ I={{\sin }^{-1}}\left( 1 \right)-{{\sin }^{-1}}\left( 1 \right) $$

$$ I=0 $$

 

Hence, the value is $$0$$.

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