Problem 10
Question
Perform the indicated integrations. $$ \int \sin ^{1 / 2} 2 z \cos ^{3} 2 z d z $$
Step-by-Step Solution
Verified Answer
The integral is \( \frac{1}{3} (\sin 2z)^{3/2} - \frac{1}{7} (\sin 2z)^{7/2} + C \).
1Step 1: Substitution to Simplify the Integral
First, let's use the substitution method to simplify the integral. Let \( u = \sin 2z \), which implies \( du = 2 \cos 2z \, dz \) or \( \frac{1}{2} \, du = \cos 2z \, dz \). The given integral becomes \( \frac{1}{2} \int u^{1/2} \cos^{2} 2z \, du \). Using the identity \( \cos^{2} 2z = 1 - \sin^{2} 2z = 1 - u^2 \), it changes to \( \frac{1}{2} \int u^{1/2} (1 - u^2) \, du \).
2Step 2: Distribute and Simplify the Integral
Now, let's distribute the \( u^{1/2} \) across the terms inside the parenthesis: \( \frac{1}{2} \int (u^{1/2} - u^{5/2}) \, du \). Split the integral: \( \frac{1}{2} \left( \int u^{1/2} \, du - \int u^{5/2} \, du \right) \).
3Step 3: Integrate Each Term Separately
Now, let's integrate each term separately:- For \( \int u^{1/2} \, du \), use the power rule: \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \). Thus, \( \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \).- For \( \int u^{5/2} \, du \), similarly: \( \int u^{5/2} \, du = \frac{u^{7/2}}{7/2} = \frac{2}{7} u^{7/2} \).Substitute these back into the integral: \( \frac{1}{2} \left( \frac{2}{3} u^{3/2} - \frac{2}{7} u^{7/2} \right) \).
4Step 4: Simplify and Substitute Back
Simplify the expression by multiplying by the \( \frac{1}{2} \) factor: \( \frac{1}{3} u^{3/2} - \frac{1}{7} u^{7/2} \). Substitute back the value of \( u \) from Step 1, which is \( u = \sin 2z \): \( \frac{1}{3} (\sin 2z)^{3/2} - \frac{1}{7} (\sin 2z)^{7/2} + C \), where \( C \) is the integration constant.
Key Concepts
Substitution MethodTrigonometric IntegralsIntegration by Parts
Substitution Method
The substitution method is a powerful technique used in calculus integration to simplify complex integrals. It's like changing the variable to make the problem easier to solve.
In this exercise, we started with the integral \( \int \sin^{1/2} 2z \cos^{3} 2z \, dz \). To simplify it, choose a substitution that minimizes the complexity, such as letting \( u = \sin 2z \). This turns the integral into terms of \( u \) instead of \( z \).
When choosing a substitution, ensure the derivative of your new variable is present in the integral. Here, \( du = 2 \cos 2z \, dz \) fits perfectly because the term \( \cos^{3} 2z \) provides a suitable match for integration. Thus, the integral becomes
In this exercise, we started with the integral \( \int \sin^{1/2} 2z \cos^{3} 2z \, dz \). To simplify it, choose a substitution that minimizes the complexity, such as letting \( u = \sin 2z \). This turns the integral into terms of \( u \) instead of \( z \).
When choosing a substitution, ensure the derivative of your new variable is present in the integral. Here, \( du = 2 \cos 2z \, dz \) fits perfectly because the term \( \cos^{3} 2z \) provides a suitable match for integration. Thus, the integral becomes
- \( \frac{1}{2} \int u^{1/2} (1 - u^2) \, du \)
Trigonometric Integrals
Trigonometric integrals often involve the integration of powers of sine and cosine functions. These are common in calculus problems.
They can be challenging, but using trigonometric identities can simplify the process.
In this exercise, the original integral \( \int \sin^{1/2} 2z \cos^{3} 2z \, dz \) involves trig functions. By employing the identity \( \cos^2 2z = 1 - \sin^2 2z \), we made the substitution to rewrite it in terms of \( u \).
This step allowed us to handle the integral using algebraic expressions instead of oscillating functions.
They can be challenging, but using trigonometric identities can simplify the process.
In this exercise, the original integral \( \int \sin^{1/2} 2z \cos^{3} 2z \, dz \) involves trig functions. By employing the identity \( \cos^2 2z = 1 - \sin^2 2z \), we made the substitution to rewrite it in terms of \( u \).
This step allowed us to handle the integral using algebraic expressions instead of oscillating functions.
- Always look for opportunities to apply trigonometric identities when dealing with integrals involving sine and cosine.
Integration by Parts
Integration by parts is a method used when standard integration techniques do not easily apply. It relies on transforming the integral of a product of functions into an easier problem.
While we didn't use integration by parts in every step of this solution, it is an important technique to understand.
In this exercise, breaking down the integral into smaller parts facilitated solving through substitution. Remember, integration by parts can simplify otherwise complicated integrals, particularly when direct substitution or straightforward antiderivatives are not evident.
While we didn't use integration by parts in every step of this solution, it is an important technique to understand.
- For any integral of the form \( \int u \, dv \), you can use the formula \( \int u \, dv = uv - \int v \, du \)
In this exercise, breaking down the integral into smaller parts facilitated solving through substitution. Remember, integration by parts can simplify otherwise complicated integrals, particularly when direct substitution or straightforward antiderivatives are not evident.
Other exercises in this chapter
Problem 10
Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{2 x^{2}-x-20}{x^{2}+x-6} d x\)
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Evaluate the given integral. $$ \int_{3}^{4} \frac{1}{t-\sqrt{2 t}} d t $$
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Perform the indicated integrations. $$ \int_{0}^{4} \frac{5}{\sqrt{2 t+1}} d t $$
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$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int t \sqrt[3]{2 t+7} d t $$
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