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Pattern - 14 of Indefinite Integration

February 18, 2021
Category: Pattern - 14 of Indefinite Integration

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⋆ Pattern – 14 ( Some special formulas )

$1)\ \int\sqrt{x^2+a^2}\ dx=\frac{x}{2}\sqrt{x^2+a^2}+\ \frac{a^2}{2}\log{\left|x+\ \sqrt{x^2+a^2}\right|}+c$

$2)\ \int\sqrt{x^2-a^2}\ dx=\frac{x}{2}\sqrt{x^2-a^2}-\ \frac{a^2}{2}\log{\left|x+\ \sqrt{x^2-a^2}\right|}+c$

$3)\ \int\sqrt{a^2-x^2}=\ \frac{x}{2}\ \sqrt{a^2-x^2}+\ \frac{a^2}{2}\sin^{-1}{\frac{x}{a}}+c$

$1)\ I=\ \int\sqrt{4-3x-2x^2}\ dx$

$=\ \int\sqrt{4-\ \left(2x^2+3x\right)}\ dx$

$=\ \int\sqrt{4-\ \left[2x^2+3x+\frac{9}{8}\right]}\ dx$

$=\ \int{\sqrt{\frac{41}{8}-\ \left(\sqrt2x+\ \frac{3}{2\ \sqrt2}\right)^2}\ }dx$

$=\ \int\sqrt{\left(\frac{\sqrt{41}}{2\sqrt2}\right)^2-\ \left(\sqrt2x+\ \frac{3}{2\sqrt2}\right)^2}\ dx$

$=\frac{1}{\sqrt2}\left[\frac{\left(\sqrt2x+ \frac{3}{2\sqrt2}\right)}{2}\ \sqrt{4-3x-2x^2}+\ \frac{41}{8\times2}\sin^{-1}{\left(\frac{\sqrt2x+ \frac{3}{2\sqrt2}}{\frac{\sqrt{41}}{2\sqrt2}}\right)}\right]+c$

$=\ \frac{1}{\sqrt2}\ \left[\frac{\left(4x+3\right)}{4\sqrt2}\ \ \sqrt{4-3x-2x^2}+\frac{41}{16}\sin^{-1}{\left(\frac{4x+3}{\sqrt{41}}\right)}\right]+c$