Question

Evaluate

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

Answer 1

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

Complete the square

$=\int \sqrt{\frac{9}{8}-{(\sqrt{2}x-\frac{1}{2^{3/2}})}^2}dx$

$=\frac{1}{2^{3/2}}\int \sqrt{9-(4x-1{)}^2}dx$

Substitute $u=4x-1$ $\to dx=\frac{1}{4}du$

$=\frac{1}{2^{3/2}}\frac{1}{4}\int \sqrt{9-u^2}du$

$=\frac{1}{2^{3/2}}\cdot \frac{3u\sqrt{1-u^2/9}}{8}+\frac{9{\sin }^{-1}\left(4/3\right))}{8}$

$=\frac{1}{2^{3/2}}\cdot \frac{9{\sin }^{-1}\left(\frac{4x-1}{3}\right)}{8}+\frac{3(4x-1)\sqrt{\frac{1-(4x-1{)}^2}{9}}}{8}$

$=\frac{1}{2^{3/2}}[\frac{9{\sin }^{-1}\left(\frac{4x-1}{3}\right)}{2^{4/2}}+\frac{3(4x-1)\sqrt{1-\frac{(4x-1{)}^2}{9}}}{2^{9/2}}]$.