Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $${\int }_0^{\pi /2}{\sin }^2\theta \cos \theta d\theta$$

Short answer

Question: Evaluate the definite integral: $${\int }_0^{\pi /2}{\sin }^2\theta \cos \theta d\theta$$ Answer: $$\frac{1}{3}$$

Step-by-step solution

Define the substitution

Let's make the substitution $$u=\sin \theta$$.

Find du in terms of dθ

To find $$du$$ in terms of $$d\theta$$, we differentiate $$u$$ with respect to $$\theta$$: $$\frac{du}{d\theta }=\frac{d(\sin \theta )}{d\theta }$$ We know that $$\frac{d(\sin \theta )}{d\theta }=\cos \theta$$. Therefore, we have $$du=\cos \theta d\theta$$

Change the limits of integration

We need to change the limits of integration from $$\theta$$ to $$u$$. Notice that the limits given are $$0$$ and $$\frac{\pi }{2}$$. To determine the corresponding limits for $$u$$, we simply substitute these values into the $$u$$ expression: When $$\theta =0$$, we have $$u=\sin (0)=0$$. So our lower limit for the $$u$$ integral is $$0$$. When $$\theta =\frac{\pi }{2}$$, we have $$u=\sin (\frac{\pi }{2})=1$$. So our upper limit for the $$u$$ integral is $$1$$.

Rewrite the integral in terms of u

Now that we have our substitution and new limits, rewrite the integral in terms of $$u$$ and $$du$$: $${\int }_0^{\pi /2}{\sin }^2\theta \cos \theta d\theta$$ $$={\int }_0^1{u}^{2}du$$

Evaluate the integral

Now evaluate the definite integral: $${\int }_0^1{u}^{2}du=\frac{1}{3}{u}^3{|}_0^1$$ $$=\frac{1}{3}(1{)}^3-\frac{1}{3}(0{)}^3$$ $$=\frac{1}{3}-0$$ $$=\frac{1}{3}$$ So the final result is: $${\int }_0^{\pi /2}{\sin }^2\theta \cos \theta d\theta =\frac{1}{3}$$

Key concepts

Substitution Method

The substitution method is a powerful technique in calculus for evaluating integrals, especially when the integral is cumbersome or complex in its initial form. In this method, you transform the original variable into a new one, to simplify the integral. This is particularly useful for trigonometric integrals like the one we are tackling here.

To apply the substitution method:

• Identify a function inside the integral that can be set as a new variable (substitute $u$ for $\sin \theta$ in our example).
• Calculate the differential of this function (e.g., $du=\cos \theta d\theta$).
• Replace both the variable and its differential within the integral.

By executing these steps, the integral is often converted into a simpler form, allowing for straightforward evaluation. It's like turning a complicated puzzle into a simpler one that is easier to solve.

Trigonometric Integrals

Trigonometric integrals involve functions such as sine and cosine and can often look intimidating. However, with methods like substitution, they become more manageable. Let’s take a closer look at how to handle these kinds of integrals.

In the exercise, we're dealing with ${\sin }^2\theta \cos \theta$, which is a product of trigonometric functions. Using substitution, we treated $\sin \theta$ as a single entity ($u$), making $\cos \theta d\theta$ the corresponding $du$. This step simplifies the expression significantly.

When dealing with trigonometric integrals involving powers of sine and cosine:

• If possible, use substitution to convert the trigonometric integral into a polynomial expression.
• Leverage trigonometric identities where appropriate to simplify the process.
• Transforming the integral table or common integral forms can save time and effort.

Using these strategies, trigonometric integrals become much more approachable and less daunting.

Limits of Integration

When changing the variable of an integral, as we do in substitution, it's crucial to correctly transform the limits of integration. The limits define the region over which you are integrating, so altering them properly ensures the result is still accurate.

In our example, the original limits were $0$ and $\pi /2$ with respect to $\theta$. To translate these:

• Substitute the lower limit into the substitution equation $u=\sin (\theta )$: $\sin (0)=0$, making the new lower limit $u=0$.
• For the upper limit, $\sin (\pi /2)=1$, making the new upper limit $u=1$.

Correctly managing the limits of integration is essential in substitution to maintain the integrity of the definite integral.

Definite Integral Evaluation

Finally, evaluating the definite integral is the part where all previous work culminates. With the correctly substituted variables and transformed limits, the integral often becomes very straightforward.

In our problem, after substitution, the integral ${\int }_0^1{u}^{2}du$ is quite simple. To evaluate:

• Find the antiderivative of the new integral: $\frac{1}{3}{u}^3$.
• Apply the new limits: ${\left[\frac{1}{3}{u}^3\right]}_0^1$.
• Compute the result: subtract the value at the lower limit from that at the upper one. This yields $\frac{1}{3}$.

By following these steps, the target definite integral is evaluated efficiently and accurately, resulting in $\frac{1}{3}$ for this specific exercise. The methodical approach ensures every integral solved leaves no room for error.