To solve this integral using the given substitution, let's work through it step-by-step.

Given Problem

Evaluate the integral: $\int \frac{28a^3 \, da}{\sqrt{3 - a^4}}$ with the substitution: $u = 3 - a^4$

Solution Steps

1. Differentiate $u$ with respect to $a$: $u = 3 - a^4 \Rightarrow \frac{du}{da} = -4a^3$ This implies: $du = -4a^3 \, da \quad \Rightarrow \quad a^3 \, da = -\frac{du}{4}$

2. Substitute into the integral: Substitute $u = 3 - a^4$ and $a^3 \, da = -\frac{du}{4}$ into the integral: $\int \frac{28a^3 \, da}{\sqrt{3 - a^4}} = \int \frac{28 \cdot -\frac{du}{4}}{\sqrt{u}}$ Simplifying this, we get: $= -7 \int \frac{du}{\sqrt{u}}$

3. Evaluate the integral in terms of $u$: The integral now becomes: $-7 \int u^{-\frac{1}{2}} \, du$ Integrate $u^{-\frac{1}{2}}$: $-7 \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = -14 u^{\frac{1}{2}} = -14 \sqrt{u}$

4. Substitute back $u = 3 - a^4$: $-14 \sqrt{u} = -14 \sqrt{3 - a^4}$

5. Add the constant of integration $C$: $-14 \sqrt{3 - a^4} + C$

Conclusion

The answer is: $\boxed{-14 \sqrt{3 - a^4} + C}$

Verification with Options

Based on the provided choices:

• Option D: $-14 \sqrt{3 - a^4} + C$ is correct.

Would you like a deeper explanation, or do you have any questions?

Related Questions

1. How do you handle substitution when the integrand contains a complex radical?
2. What are some common substitutions for integrals involving square roots?
3. How does substitution affect the bounds of definite integrals?
4. What techniques can you use if substitution doesn’t simplify the integral?
5. How can integration by parts be combined with substitution?

Tip

Always verify the substitution by differentiating to ensure it matches the differential in the original integral.