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Evaluate the integral of \int\frac{1}{5\sqrt{25-4x^{2}}}dx

Option: 1

\frac{1}{4x}sec^{-1}\frac{5}{2x}+C


Option: 2

-\frac{1}{4x}sec^{-1}\frac{5}{2x}+C


Option: 3

\frac{1}{5}sec^{-1}\frac{4x}{5}+C


Option: 4

-\frac{1}{5}sec^{-1}\frac{4x}{5}+C


Answers (1)

best_answer

Given integral,

\int\frac{1}{5\sqrt{25-4x^{2}}}dx

\int\frac{1}{5\sqrt{25-4x^{2}}}dx=\int\frac{1}{5\sqrt{5^{2}-\left ( 2x \right )^{2}}}dx

Comparing the given integral with the integral formula

\int \frac{1}{u\sqrt{u^{2}-a^{2}}}du=\frac{1}{a}sec^{-1}\frac{u}{a}+C

Therefore,

\int\frac{1}{5\sqrt{25-4x^{2}}}dx=\frac{1}{2x}sec^{-1}\frac{5}{2x}+C

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