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webwork 5 - topics 10 - 12: pr
(1 point)
use an appropriate substitution to compute the indefinite integral.
$$ int \frac { x ^ { 2 } } { sqrt { 1 - 4 x ^ { 6 } } } d x = square + c. $$
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Explanation:

Step1: Choose substitution

Let $u = 2x^3$, so $du = 6x^2 dx \implies x^2 dx = \frac{du}{6}$.

Step2: Rewrite integral

Original integral becomes $\int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{6}$.

Step3: Integrate

$\frac{1}{6} \int \frac{1}{\sqrt{1 - u^2}} du = \frac{1}{6} \arcsin(u) + C$.

Step4: Substitute back

Replace $u$ with $2x^3$: $\frac{1}{6} \arcsin(2x^3) + C$.

Answer:

$\frac{1}{6}\arcsin(2x^3)$