Math Problem Statement

Evaluate the indefinite integral: ∫ x^4/√1-x^10 dx

Solution

To evaluate the indefinite integral

x41x10dx\int \frac{x^4}{\sqrt{1 - x^{10}}} \, dx

Step-by-Step Solution:

  1. Substitution: Let u=x5u = x^5. Then: du=5x4dxor15du=x4dxdu = 5x^4 \, dx \quad \text{or} \quad \frac{1}{5} du = x^4 \, dx Also, notice that x10=(x5)2=u2x^{10} = (x^5)^2 = u^2.

  2. Rewriting the integral: Substituting x5=ux^5 = u and x4dx=15dux^4 \, dx = \frac{1}{5} du, the integral becomes: x41x10dx=15du1u2\int \frac{x^4}{\sqrt{1 - x^{10}}} \, dx = \int \frac{\frac{1}{5} du}{\sqrt{1 - u^2}}

  3. Simplifying: Factor out the constant 15\frac{1}{5}: 1511u2du\frac{1}{5} \int \frac{1}{\sqrt{1 - u^2}} \, du

  4. Recognizing the standard integral: The integral 11u2du\int \frac{1}{\sqrt{1 - u^2}} \, du is a standard form, which equals arcsin(u)\arcsin(u).

    Thus: 1511u2du=15arcsin(u)+C\frac{1}{5} \int \frac{1}{\sqrt{1 - u^2}} \, du = \frac{1}{5} \arcsin(u) + C

  5. Back-substitution: Recall that u=x5u = x^5, so: 15arcsin(u)+C=15arcsin(x5)+C\frac{1}{5} \arcsin(u) + C = \frac{1}{5} \arcsin(x^5) + C

Final Answer:

x41x10dx=15arcsin(x5)+C\int \frac{x^4}{\sqrt{1 - x^{10}}} \, dx = \frac{1}{5} \arcsin(x^5) + C


Would you like further clarification on any of the steps?

Related Questions:

  1. How is the substitution method applied in integral problems?
  2. What is the derivation of the formula 11x2dx=arcsin(x)+C\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C?
  3. How does substitution help simplify integrals with powers and roots?
  4. Can you provide a geometrical interpretation of arcsin(x5)\arcsin(x^5)?
  5. What are other techniques for solving integrals involving square roots?

Tip:

When encountering integrals with 1xn\sqrt{1 - x^n}, consider substitution to simplify powers or recognize standard trigonometric forms.

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Math Problem Analysis

Mathematical Concepts

Integration
Substitution method
Trigonometric integrals

Formulas

∫ 1/√(1 - u^2) du = arcsin(u) + C
Substitution: Let u = g(x), then du = g'(x) dx

Theorems

Fundamental theorem of calculus
Standard integrals involving square roots

Suitable Grade Level

Grades 11-12