Solve the following systems of equations:

Question:

$\int \frac{d x}{\sqrt{16-9 x^{2}}}$

Solution:

Let, $\mathrm{I}=\int \frac{d x}{\sqrt{16-9 x^{2}}}$

$=\frac{1}{3} \int \frac{d x}{\sqrt{\frac{16}{9}-x^{2}}}=\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{2}\right)^{2}-x^{2}}}$

$=\frac{1}{3} \sin ^{-1} \frac{x}{4 / 3}+C\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C\right]$

$=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$

Therefore, $\mathrm{I}=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$.

 

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