Problem 13
Question
In Problems 1-54, perform the indicated integrations. \(\int \frac{\sin \sqrt{t}}{\sqrt{t}} d t\)
Step-by-Step Solution
Verified Answer
The integral is \(-2 \cos(\sqrt{t}) + C\).
1Step 1: Identify the Integration Technique
We need to solve the integral \( \int \frac{\sin \sqrt{t}}{\sqrt{t}} dt \). Notice that the integrand involves a function of \( \sqrt{t} \), which suggests a substitution method might help here.
2Step 2: Make a Substitution
Set \( u = \sqrt{t} \), so \( t = u^2 \). This implies \( \frac{dt}{du} = 2u \), or \( dt = 2u \, du \).
3Step 3: Substitute and Simplify the Integral
Substitute \( u \) and \( dt \) into the integral:\[ \int \frac{\sin \sqrt{t}}{\sqrt{t}} dt = \int \frac{\sin u}{u} \cdot 2u \, du = 2 \int \sin u \, du. \]The \( u \) terms cancel, simplifying the integral to \( 2 \int \sin u \, du \).
4Step 4: Integrate Using Known Formulas
Recall that the integral of \( \sin u \) is \( -\cos u \). Therefore, integrate:\[ 2 \int \sin u \, du = 2(-\cos u) = -2 \cos u + C, \]where \( C \) is the constant of integration.
5Step 5: Substitute Back to Original Variable
Replace \( u \) with the original variable \( \sqrt{t} \) to finalize the integration:\[-2 \cos u + C = -2 \cos(\sqrt{t}) + C.\]
Key Concepts
Substitution MethodIntegral of Sine FunctionConstant of Integration
Substitution Method
The substitution method in calculus is like a nifty shortcut that simplifies complex integrals, especially when they involve composite functions. The aim is to transform the integral into a simpler form, making it more manageable to solve. Imagine having an intricate puzzle that seems difficult at first glance. The substitution method allows you to adjust the pieces so they all fit seamlessly, revealing the final picture more clearly.
In our example, the integral \( \int \frac{\sin \sqrt{t}}{\sqrt{t}} dt \) involves \( \sqrt{t} \), a function of \( t \), suggesting a substitution could simplify it. A good rule of thumb when approaching such integrals is to look for a substitution that clears the complexity. Here, setting \( u = \sqrt{t} \) transforms \( t \) into \( u^2 \).
Once we identify our substitution, we also need to express \( dt \) in terms of \( du \). By differentiating \( u = \sqrt{t} \), we find \( dt = 2u \, du \). Substituting these into the integral, the new expression becomes \( 2 \int \sin u \, du \). This technique works wonders by stripping down the integral to a more recognizable form, thus easing our calculation efforts.
In our example, the integral \( \int \frac{\sin \sqrt{t}}{\sqrt{t}} dt \) involves \( \sqrt{t} \), a function of \( t \), suggesting a substitution could simplify it. A good rule of thumb when approaching such integrals is to look for a substitution that clears the complexity. Here, setting \( u = \sqrt{t} \) transforms \( t \) into \( u^2 \).
Once we identify our substitution, we also need to express \( dt \) in terms of \( du \). By differentiating \( u = \sqrt{t} \), we find \( dt = 2u \, du \). Substituting these into the integral, the new expression becomes \( 2 \int \sin u \, du \). This technique works wonders by stripping down the integral to a more recognizable form, thus easing our calculation efforts.
Integral of Sine Function
The integral of the sine function is a common element in calculus, yet it's a fundamental stepping stone to understanding more complex problems. For many students, recognizing how to handle these basic integrals is key to tackling larger calculus challenges.
In our transformed integral \( 2 \int \sin u \, du \), we deal with an easily recognizable piece: the sine function. The integral of \( \sin u \) is \(-\cos u\). That's a universal rule in calculus and serves as a base for many integral calculations.
To integrate \( 2 \int \sin u \, du \), we use the known result \(-\cos u\). Each integral of this sort will have similar steps:
In our transformed integral \( 2 \int \sin u \, du \), we deal with an easily recognizable piece: the sine function. The integral of \( \sin u \) is \(-\cos u\). That's a universal rule in calculus and serves as a base for many integral calculations.
To integrate \( 2 \int \sin u \, du \), we use the known result \(-\cos u\). Each integral of this sort will have similar steps:
- Recognize the sine function within the integral.
- Apply the rule for integrating sine, which gives \(-\cos u\).
- Multiply by outside constants accordingly, in this case, 2, resulting in \(-2 \cos u\).
Constant of Integration
In the world of indefinite integrals, the constant of integration \( C \) plays a crucial role. It might seem like a small detail, but it's a vital part of the solution. When performing integration, especially for indefinite integrals, we're essentially dealing with families of functions that only differ by a constant shift along the y-axis.
After we solve our example integral, we arrive at the expression \(-2 \cos(\sqrt{t}) + C\). Here, \( C \) represents an unknown constant that acknowledges there are multiple possible antiderivatives for a given function. Each one is reached by adding a constant to the result of the definite integration.
It's important never to forget \( C \) in your final answers when dealing with indefinite integrals. It serves to remind us of the continuous nature of integrals and their solutions, highlighting that while we can pinpoint a class of functions that can be solutions, the precise function is determined once additional information is provided, such as initial conditions in practical problems. Remember:
After we solve our example integral, we arrive at the expression \(-2 \cos(\sqrt{t}) + C\). Here, \( C \) represents an unknown constant that acknowledges there are multiple possible antiderivatives for a given function. Each one is reached by adding a constant to the result of the definite integration.
It's important never to forget \( C \) in your final answers when dealing with indefinite integrals. It serves to remind us of the continuous nature of integrals and their solutions, highlighting that while we can pinpoint a class of functions that can be solutions, the precise function is determined once additional information is provided, such as initial conditions in practical problems. Remember:
- Always include \( C \) with indefinite integrals.
- The constant ensures that solutions account for all possible vertical shifts.
- It embodies the generality of antiderivatives, applicable across multiple contexts.
Other exercises in this chapter
Problem 12
In Problems 1-14, solve each differential equation. $$ y^{\prime}=e^{2 x}-3 y ; y=1 \text { when } x=0 $$
View solution Problem 12
In Problems 1-36, use integration by parts to evaluate each integral. $$ \int \ln \left(7 x^{5}\right) d x $$
View solution Problem 13
In Problems 11-16, use Euler's Method with \(h=0.2\) to approximate the solution over the indicated interval. $$ y^{\prime}=x, y(0)=0,[0,1] $$
View solution Problem 13
In Problems 1-28, perform the indicated integrations. \(\int \sin 4 y \cos 5 y d y\)
View solution