SOLUTION 3: $\ \$ To integrate $\displaystyle{ \int { 1 \over (1-x^2)^{3/2} } \ dx }$ use the trig substitution $$x = \sin \theta$$ so that $$dx = \cos \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $x$, getting $$\displaystyle { \int \frac{1}{(1-x^{2})^{3/2}} \ dx = \int \frac{1}{(1-\sin^{2} \theta)^{3/2}} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{(\cos^{2} \theta)^{3/2}} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{\cos^{3} \theta} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{\cos^{2} \theta} \ d \theta }$$ $$= \displaystyle { \int \sec^{2} \theta \ d \theta }$$ $$= \tan \theta + C$$ $\Big($ We need to write our final answer in terms of $x$.

Since $x = \sin \theta$ it follows that $$\sin \theta = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over hypotenuse }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x)^2 = (1)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{1-x^2} \ \ \longrightarrow$$ $$\tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ x \over \sqrt{1-x^2} } . \Big)$$ $$= \displaystyle{ x \over \sqrt{1-x^2} } + C$$

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