Mathematics

Evaluate the following integral:
$$\displaystyle \int { \cfrac { 1 }{ \sqrt { 5-4x-2{ x }^{ 2 } }  }  } dx$$


SOLUTION
$$\displaystyle\int{\dfrac{dx}{\sqrt{5-4x-2{x}^{2}}}}$$

Consider $$5-4x-2{x}^{2}=-\left(2{x}^{2}+4x-5\right)$$

$$=-2\left({x}^{2}+2x-\dfrac{5}{2}\right)$$

$$=-2\left({x}^{2}+2x+1-1-\dfrac{5}{2}\right)$$

$$=-2\left({\left(x+1\right)}^{2}+\dfrac{-2-5}{2}\right)$$

$$=-2\left({\left(x+1\right)}^{2}+\dfrac{-7}{2}\right)$$

$$=2\left(\dfrac{7}{2}-{\left(x+1\right)}^{2}+\right)$$

$$=2\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)$$

$$\displaystyle\int{\dfrac{dx}{\sqrt{5-4x-2{x}^{2}}}}$$

$$=\displaystyle\int{\dfrac{dx}{\sqrt{2\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)}}}$$

$$=\dfrac{1}{\sqrt{2}}\displaystyle\int{\dfrac{dx}{\sqrt{\left({\left(\dfrac{7}{2}\right)}^{2}-{\left(x+1\right)}^{2}+\right)}}}$$

We know that $$\displaystyle\int{\dfrac{dx}{\sqrt{{a}^{2}-{x}^{2}}}}={\sin}^{-1}{\dfrac{x}{a}}+c$$

Replace $$x\rightarrow\,x\rightarrow\,x+1$$ and $$a\rightarrow\,\sqrt{\dfrac{7}{2}}$$

$$=\dfrac{1}{\sqrt{2}}{\sin}^{-1}{\dfrac{x+1}{\sqrt{\dfrac{7}{2}}}}+c$$

$$=\dfrac{1}{\sqrt{2}}{\sin}^{-1}{\sqrt{\dfrac{2}{7}}\left(x+1\right)}+c$$
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