Question

Perform the indicated integrations. $$ \int \sin ^{1 / 2} 2 z \cos ^{3} 2 z d z $$

Short answer

The integral is \( \frac{1}{3} (\sin 2z)^{3/2} - \frac{1}{7} (\sin 2z)^{7/2} + C \).

Step-by-step solution

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 \( \frac{1}{2} \int u^{1/2} (1 - u^2) \, du \).

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: \( \frac{1}{2} \left( \int u^{1/2} \, du - \int u^{5/2} \, du \right) \).

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: \( \frac{1}{2} \left( \frac{2}{3} u^{3/2} - \frac{2}{7} u^{7/2} \right) \).

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 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

• \( \frac{1}{2} \int u^{1/2} (1 - u^2) \, du \)

which is much easier to solve.

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.

• 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.

• For any integral of the form \( \int u \, dv \), you can use the formula \( \int u \, dv = uv - \int v \, du \)

where \( u \) and \( dv \) are defined based on the integral components.
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.