Problem 4
Question
$$\int_{0}^{\pi} \cos x \sin ^{2} x d x=\left.\frac{1}{3} \sin ^{3} x\right|_{0} ^{\pi}=0$$
Step-by-Step Solution
Verified Answer
The integral \( \int_{0}^{\pi} \cos x \sin^2 x \, dx \) evaluates to 0.
1Step 1: Identify the Integration Method
Observe the integrand \( \cos x \sin^2 x \). Notice that integration by substitution is a suitable method here, particularly because \( \sin x \) and its derivative, \( \cos x \, dx \), are present.
2Step 2: Choose a Substitution Variable
Set \( u = \sin x \). Then, the differential \( du = \cos x \, dx \). This substitution simplifies the integral.
3Step 3: Rewrite the Integral with Substitution
Substitute \( u \) and \( du \) into the integral. The limits of integration change according to \( x \). When \( x = 0 \), \( u = \sin(0) = 0 \). When \( x = \pi \), \( u = \sin(\pi) = 0 \). The integral becomes \( \int_{0}^{0} u^2 \, du \).
4Step 4: Evaluate the Definite Integral
Evaluate \( \int_{0}^{0} u^2 \, du \), which is \( \frac{1}{3} u^3 \bigg|_{0}^{0} \). Since both the upper and lower limits are 0, the result is 0.
5Step 5: Verify the Solution
Ensure each step follows from the previous correctly. The substitution method is valid, and the computation confirms that the original integral evaluates to 0.
Key Concepts
Integration by SubstitutionDefinite IntegralsTrigonometric Integration
Integration by Substitution
Integration by substitution is a powerful technique used when dealing with integrals. This method simplifies the process of integration by substituting a part of the integral with a new variable, making it easier to solve. The key is to identify an inner function and its derivative within the integrand.
Typically, if you see something like \(f(g(x)) \, g'(x)\) in the integrand, substitution might work well.
Typically, if you see something like \(f(g(x)) \, g'(x)\) in the integrand, substitution might work well.
- Identify the function inside the integral that could be swapped out for easier integration.
- Set this function as a new variable, say \(u\).
- Express \(du\), the derivative of \(u\), in terms of \(dx\).
Definite Integrals
Definite integrals are important in determining the actual numerical value of the area under a curve. Unlike indefinite integrals, they assign specific values to integrals over a given interval, saying essentially "this much space lies beneath the curve between these two points."
For definite integrals, limits of integration must be specified. These are the values at which you evaluate the integral to find the enclosed area. For example, when you have \(\int_a^b f(x)\,dx\), \(a\) and \(b\) are your limits of integration, the x-values that mark the beginning and end of the range you're interested in.
To solve a definite integral:
For definite integrals, limits of integration must be specified. These are the values at which you evaluate the integral to find the enclosed area. For example, when you have \(\int_a^b f(x)\,dx\), \(a\) and \(b\) are your limits of integration, the x-values that mark the beginning and end of the range you're interested in.
To solve a definite integral:
- Firstly, calculate the indefinite integral.
- Next, substitute the upper limit into this integrated function.
- Then substitute the lower limit and subtract the two results.
Trigonometric Integration
Trigonometric integration deals specifically with integrals involving trigonometric functions like sine, cosine, tangent, and so on. These integrals might look complex initially, but they can often be decomposed into simpler parts using identities and substitution.
When facing trigonometric integrals, consider:
When facing trigonometric integrals, consider:
- Utilizing trigonometric identities, such as \( an^2 x = ext{sec}^2 x - 1\) or \( ext{sin}^2 x = 1 - ext{cos}^2 x\).
- Substitution, especially when derivatives of trigonometric functions appear together, such as the pair \(\sin x\) and \(\cos x \, dx\).
- Reducing the power of trigonometric functions using identities to simplify integration.
Other exercises in this chapter
Problem 3
$$\int_{0}^{2} e^{x}\left(x e^{-x}-e^{-x}\right) d x=\int_{0}^{2}(x-1) d x=\left.\left(\frac{1}{2} x^{2}-x\right)\right|_{0} ^{2}=0$$
View solution Problem 4
$$\begin{aligned} a_{0} &=\int_{-1}^{1} f(x) d x=\int_{0}^{1} x d x=\frac{1}{2} \\ a_{n} &=\int_{-1}^{1} f(x) \cos n \pi x d x=\int_{0}^{1} x \cos n \pi x d x=\
View solution Problem 5
$$\begin{aligned} c_{i} &=\frac{2 \alpha_{i}^{2}}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \int_{0}^{2} x J_{0}\left(\alpha_{i} x\rig
View solution Problem 5
Since \(f(-x)=e^{|-x|}=e^{|x|}=f(x), f(x)\) is an even function.
View solution