Mathematics

Evaluate the following integrals:
$$\int { \sqrt { 1+x-2{ x }^{ 2 } }  } dx$$


SOLUTION
Let $$I=\displaystyle\int \sqrt{1+x-2 x^2} d x$$

        $$=\displaystyle\int \sqrt{1-2\left(-2(x)\bigg(\dfrac{1}{4}\bigg)+x^2\right)}\  d x$$

        $$=\displaystyle\int \sqrt{1+\dfrac{1}{8}-2\left(\bigg(\dfrac{1}{4}\bigg)^2-2(x)\bigg(\dfrac{1}{4}\bigg)+x^2\right)}\ d x$$

       $$=\sqrt{2}\displaystyle\int \sqrt{\dfrac{9}{16}-\bigg(x-\dfrac{1}{4}\bigg)^2}\ d x$$                                                        $$(\because (a-b)^2=a^2-2 a b+b^2)$$

Put $$t=x-\dfrac{1}{4}\implies d t=d x$$

$$I=\sqrt{2}\displaystyle\int \sqrt{\bigg(\dfrac{3}{4}\bigg)^2-t^2}\ d t$$

As we know that

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

Here $$a=\dfrac{3}{4}$$

$$\implies I=\sqrt{2}\left(\dfrac{t}{2}\sqrt{\dfrac{9}{16}-t^2}+\dfrac{9/16}{2}\text{sin}^{-1} \left(\dfrac{t}{3/4}\right)+C\right)$$

$$\implies I=\dfrac{1}{2}\bigg(x-\dfrac{1}{4}\bigg)\sqrt{\dfrac{9}{8}-2\bigg(x-\dfrac{1}{4}\bigg)^2}+\dfrac{9\sqrt{2}}{32}\text{sin}^{-1} \bigg(\dfrac{4(x-1/4)}{3}\bigg)+C$$

$$\implies I=\dfrac{1}{8}(4 x-1)\sqrt{1+x-2 x^2}+\dfrac{9\sqrt{2}}{32}\text{sin}^{-1} \bigg(\dfrac{4 x-1}{3}\bigg)+C$$

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