Solved: "Evaluate the integrals ∫−25−65dy|y|5y2−3"

bobbie71G

Answered question

2021-10-13

Evaluate the integrals

\int_{- \frac{2}{\sqrt{5}}}^{- \frac{\sqrt{6}}{\sqrt{5}}} \frac{d y}{| y | \sqrt{5 y^{2} - 3}}

Isma Jimenez

Skilled2021-10-14Added 84 answers

Given that:

\int_{- \frac{2}{\sqrt{5}}}^{- \frac{\sqrt{6}}{\sqrt{5}}} \frac{d y}{| y | \sqrt{5 y^{2} - 3}}

Solving the integral:

\int_{- \frac{2}{\sqrt{5}}}^{- \frac{\sqrt{6}}{\sqrt{5}}} \frac{d y}{| y | \sqrt{5 y^{2} - 3}}

To solve the above integral first we solve,

\int \frac{d y}{| y | \sqrt{5 y^{2} - 3}}

\Rightarrow \int \frac{d y}{| y | \sqrt{5 y^{2} - 3}} = \frac{y}{| y |} \int \frac{1}{| y | \sqrt{5 y^{2} - 3}} \times \frac{| y |}{y} d y

\Rightarrow \int \frac{d y}{| y | \sqrt{5 y^{2} - 3}} = \frac{y}{| y |} \int \frac{1}{y \sqrt{5 y^{2} - 3}} d y

Now,

\int \frac{1}{y \sqrt{5 y^{2} - 3}} d y = \int \frac{1}{t^{2} + 3} d t

\Rightarrow \int \frac{1}{y \sqrt{5 y^{2} - 3}} d y = \int \frac{1}{t^{2} + {(\sqrt{3})}^{2}} d t

Using the identity:

\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C

We get,

\int \frac{1}{y \sqrt{5 y^{2} - 3}} = \frac{y \tan^{- 1} \frac{\sqrt{5 y^{2} - 3}}{\sqrt{3}}}{\sqrt{3} | y |} + C

\int \frac{d y}{| y | \sqrt{5 y^{2} - 3}} = - \frac{y \sin^{- 1} (\frac{3}{\sqrt{15} | y |}}}{}

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