Integrate the function: frac{1}{sqrt{9-25 x^{ | vedclass

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Question

(A) Let $5x = t$. Then,$5 dx = dt$,which implies $dx = \frac{1}{5} dt$. Substituting these into the integral: $\int \frac{1}{\sqrt{9-25 x^{2}}} dx = \

Vedclass · Accepted Answer

$\frac{1}{5} \sin^{-1}(\frac{5x}{3}) + C$

Vedclass · Answer

(A) Let $5x = t$. Then,$5 dx = dt$,which implies $dx = \frac{1}{5} dt$. Substituting these into the integral: $\int \frac{1}{\sqrt{9-25 x^{2}}} dx = \int \frac{1}{\sqrt{3^{2}-(5x)^{2}}} dx$ $= \frac{1}{5} \int \frac{1}{\sqrt{3^{2}-t^{2}}} dt$ Using the standard integral formula $\int \frac{1}{\sqrt{a^{2}-x^{2}}} dx = \sin^{-1}(\frac{x}{a}) + C$: $= \frac{1}{5} \sin^{-1}(\frac{t}{3}) + C$ Substituting $t = 5x$ back: $= \frac{1}{5} \sin^{-1}(\frac{5x}{3}) + C$,where $C$ is an arbitrary constant.

Integrate the function: $\frac{1}{\sqrt{9-25 x^{2}}}$

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