Evaluate the integral:

Question:

Evaluate the integral:

$\int \sqrt{1+x-2 x^{2}} d x$

Solution:

Key points to solve the problem:

- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x)\left(\int g(x) d x\right) d x$

- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:

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

$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

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

$\therefore I=\int \sqrt{1-2\left(x^{2}-2\left(\frac{1}{4}\right) x\right)} d x=\int \sqrt{1-2\left(x^{2}-2\left(\frac{1}{4}\right) x+\left(\frac{1}{4}\right)^{2}\right)+2\left(\frac{1}{4}\right)^{2}} d x$

Using $a^{2}-2 a b+b^{2}=(a-b)^{2}$

We have:

$I=\int \sqrt{\frac{9}{8}-2\left(x-\frac{1}{4}\right)^{2}} d x=\int \sqrt{2} \sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}} d x$

As I match with the form: $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$

$\therefore I=\sqrt{2}\left\{\frac{x-\frac{1}{4}}{2} \sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}}+\frac{\frac{9}{16}}{2} \sin ^{-1}\left(\frac{x-\frac{1}{4}}{\frac{2}{4}}\right)\right\}+C$

$\Rightarrow I=\frac{1}{8}(4 x-1) \sqrt{2\left\{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}\right\}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}\left(\frac{4 x-1}{3}\right)+C$

$\Rightarrow I=\frac{1}{8}(4 x-1) \sqrt{1+x-2 x^{2}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}\left(\frac{4 x-1}{3}\right)+C$

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