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Magazine Chapter 7: Problem 39
Perform the indicated integrations. \(\int \frac{y}{\sqrt{16-9 y^{4}}} d y\)
Short Answer
Expert verified
The integral evaluates to
\(-\frac{\sqrt{16 - 9y^4}}{3} + C\).
Step by step solution
01
Identify the Substitution Needed
We notice that the integrand has a term of the form \( \sqrt{16 - 9y^4} \). This suggests a substitution to simplify the expression under the square root. Let's use \( y^2 = \frac{4}{3} \sin(\theta) \). This substitution is motivated by the trigonometric identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), aiming to simplify the square root expression.
02
Apply the Substitution
With \( y^2 = \frac{4}{3} \sin(\theta) \), consequently \( y = \sqrt{\frac{4}{3}} \sin(\theta) \). Then, the derivative is \( dy = \sqrt{\frac{4}{3}} \cos(\theta) d\theta \). Substitute these into the integral to get\[\int \frac{\sqrt{\frac{4}{3}} \sin(\theta)}{\sqrt{16 - 9 (\frac{4}{3} \sin(\theta))^2}} \sqrt{\frac{4}{3}} \cos(\theta) d\theta\]
03
Simplify the Integral
The term under the square root becomes \(16 - 12 \sin^2(\theta)\). Using the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\), this simplifies to \[16 \cos^2(\theta)\]. Thus, \( \sqrt{16 \cos^2(\theta)} = 4 \cos(\theta)\). Substitute this back into the integral:\[\int \frac{\frac{4}{3} \sin(\theta) \cos(\theta)}{4 \cos(\theta)} d\theta = \frac{4}{3} \int \sin(\theta) d\theta\]
04
Integrate with Respect to Theta
The integral \( \int \sin(\theta) d\theta \) can be computed as \(-\cos(\theta) + C\). Hence, our solution becomes \[\frac{4}{3} (-\cos(\theta)) + C\]where \(C\) is the constant of integration.
05
Back Substitute to Original Variable
From the substitution \( y^2 = \frac{4}{3} \sin(\theta) \), we have \( \sin(\theta) = \frac{3}{4}y^2 \). Therefore, \( \cos(\theta) = \sqrt{1 - (\frac{3}{4}y^2)^2} = \sqrt{1 - \frac{9}{16}y^4} = \frac{\sqrt{16 - 9y^4}}{4}\). Substitute back to find the solution in terms of \(y\):\[-\frac{4}{3} \cdot \frac{\sqrt{16 - 9y^4}}{4} + C = -\frac{\sqrt{16 - 9y^4}}{3} + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful tool in calculus used to simplify integrals by changing variables. In this exercise, we notice that the integrand has a square root with a complex expression:
- We looked at \[x = \sqrt{16 - 9y^4}\], and decided to simplify it using trigonometric substitutions.
- This method involves expressing the original variable in terms of a new variable (here, we used an angle \( \theta \)).
- We set \( y^2 = \frac{4}{3} \sin(\theta) \) to leverage the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), which simplifies the square root.
Trigonometric Integration
Trigonometric integration is a technique often used when dealing with expressions involving trigonometric functions. In our problem, the integral became dependent on trigonometric functions after substitution. Here鈥檚 how we handled it:
- After substituting \( y^2 = \frac{4}{3} \sin(\theta) \), we transformed the integration into one involving \( \sin(\theta) \) and \( \cos(\theta) \).
- The integral was simplified to \( \frac{4}{3} \int \sin(\theta) d\theta \).
- This integral is straightforward and integrates to \( -\cos(\theta) + C \).
Square Root Simplification
Simplifying expressions under a square root is often necessary to make an integral manageable. In this exercise, the square root expression \( \sqrt{16 - 9y^4} \) play a central role. Let鈥檚 break down our approach:
- We used substitution initially to transform \( 16 - 9y^4 \) into an expression involving a trigonometric function.
- Through substitution and trigonometric identities, \( \sqrt{16 - 9(\frac{4}{3} \sin(\theta))^2} \) simplified to \( 4 \cos(\theta) \).
- The key was recognizing the structure that matched with trigonometric identities to enable this simplification.
