Get comprehensive answers to your questions with the help of IDNLearn.com's community. Our platform is designed to provide trustworthy and thorough answers to any questions you may have.
Sagot :
Certainly! Let's evaluate each integral step-by-step:
(a) [tex]\(\int \cos^3 \theta \, d\theta\)[/tex]
To integrate [tex]\(\cos^3 \theta\)[/tex], we can use a method known as reduction formulas or trigonometric identities.
First, we use the trigonometric identity for the cosine function:
[tex]\[ \cos^3 \theta = \cos \theta \cdot \cos^2 \theta = \cos \theta \cdot (1 - \sin^2 \theta) \][/tex]
Thus, rewriting the integral:
[tex]\[ \int \cos^3 \theta \, d\theta = \int \cos \theta \cdot (1 - \sin^2 \theta) \, d\theta \][/tex]
Now, let [tex]\(u = \sin \theta\)[/tex]. Then, [tex]\(du = \cos \theta \, d\theta\)[/tex], and the integral becomes:
[tex]\[ \int \cos \theta \cdot (1 - \sin^2 \theta) \, d\theta = \int (1 - u^2) \, du \][/tex]
Integrate each term separately:
[tex]\[ \int 1 \, du - \int u^2 \, du = u - \frac{u^3}{3} + C \][/tex]
Substituting back [tex]\(u = \sin \theta\)[/tex]:
[tex]\[ \sin \theta - \frac{\sin^3 \theta}{3} + C \][/tex]
So, the antiderivative is:
[tex]\[ -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
(b) [tex]\(\int \frac{\ln \left(x^2\right)}{x^2} \, dx\)[/tex]
We start by simplifying the integrand. Recall that:
[tex]\[ \ln(x^2) = 2 \ln(x) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{2 \ln(x)}{x^2} \, dx \][/tex]
Rewrite the integral by taking the constant out:
[tex]\[ 2 \int \frac{\ln(x)}{x^2} \, dx \][/tex]
Now, use integration by parts. Let:
[tex]\[ u = \ln(x) \quad \text{and} \quad dv = \frac{1}{x^2} \, dx \][/tex]
Then, [tex]\(du = \frac{1}{x} \, dx\)[/tex] and integrating [tex]\(dv\)[/tex] gives us [tex]\(v = -\frac{1}{x}\)[/tex].
Using the integration by parts formula, [tex]\(\int u \, dv = uv - \int v \, du\)[/tex], we get:
[tex]\[ 2 \left[ \ln(x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) \, dx \right] \][/tex]
Simplify and solve the remaining integral:
[tex]\[ 2 \left[ -\frac{\ln(x)}{x} + \int \frac{1}{x^2} \, dx \right] \][/tex]
The remaining integral is straightforward:
[tex]\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \][/tex]
Putting it all together:
[tex]\[ 2 \left( -\frac{\ln(x)}{x} - \frac{1}{x} \right) = -\frac{2 \ln(x)}{x} - \frac{2}{x} + C' \][/tex]
Thus, the antiderivative is:
[tex]\[ -\frac{\ln(x^2)}{x} - \frac{2}{x} + C \][/tex]
Putting it all together:
[tex]\[ -\frac{\log(x^2)}{x} - \frac{2}{x} + C \][/tex]
Hence, based on the simplifications and results:
The integral results are:
[tex]\[ (a) \, \int \cos^3 \theta \, d\theta = -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
[tex]\[ (b) \, \int \frac{\ln \left(x^2\right)}{x^2} \, dx = -\frac{\log(x^2)}{x} - \frac{2}{x} + C \][/tex]
(a) [tex]\(\int \cos^3 \theta \, d\theta\)[/tex]
To integrate [tex]\(\cos^3 \theta\)[/tex], we can use a method known as reduction formulas or trigonometric identities.
First, we use the trigonometric identity for the cosine function:
[tex]\[ \cos^3 \theta = \cos \theta \cdot \cos^2 \theta = \cos \theta \cdot (1 - \sin^2 \theta) \][/tex]
Thus, rewriting the integral:
[tex]\[ \int \cos^3 \theta \, d\theta = \int \cos \theta \cdot (1 - \sin^2 \theta) \, d\theta \][/tex]
Now, let [tex]\(u = \sin \theta\)[/tex]. Then, [tex]\(du = \cos \theta \, d\theta\)[/tex], and the integral becomes:
[tex]\[ \int \cos \theta \cdot (1 - \sin^2 \theta) \, d\theta = \int (1 - u^2) \, du \][/tex]
Integrate each term separately:
[tex]\[ \int 1 \, du - \int u^2 \, du = u - \frac{u^3}{3} + C \][/tex]
Substituting back [tex]\(u = \sin \theta\)[/tex]:
[tex]\[ \sin \theta - \frac{\sin^3 \theta}{3} + C \][/tex]
So, the antiderivative is:
[tex]\[ -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
(b) [tex]\(\int \frac{\ln \left(x^2\right)}{x^2} \, dx\)[/tex]
We start by simplifying the integrand. Recall that:
[tex]\[ \ln(x^2) = 2 \ln(x) \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{2 \ln(x)}{x^2} \, dx \][/tex]
Rewrite the integral by taking the constant out:
[tex]\[ 2 \int \frac{\ln(x)}{x^2} \, dx \][/tex]
Now, use integration by parts. Let:
[tex]\[ u = \ln(x) \quad \text{and} \quad dv = \frac{1}{x^2} \, dx \][/tex]
Then, [tex]\(du = \frac{1}{x} \, dx\)[/tex] and integrating [tex]\(dv\)[/tex] gives us [tex]\(v = -\frac{1}{x}\)[/tex].
Using the integration by parts formula, [tex]\(\int u \, dv = uv - \int v \, du\)[/tex], we get:
[tex]\[ 2 \left[ \ln(x) \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) \, dx \right] \][/tex]
Simplify and solve the remaining integral:
[tex]\[ 2 \left[ -\frac{\ln(x)}{x} + \int \frac{1}{x^2} \, dx \right] \][/tex]
The remaining integral is straightforward:
[tex]\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \][/tex]
Putting it all together:
[tex]\[ 2 \left( -\frac{\ln(x)}{x} - \frac{1}{x} \right) = -\frac{2 \ln(x)}{x} - \frac{2}{x} + C' \][/tex]
Thus, the antiderivative is:
[tex]\[ -\frac{\ln(x^2)}{x} - \frac{2}{x} + C \][/tex]
Putting it all together:
[tex]\[ -\frac{\log(x^2)}{x} - \frac{2}{x} + C \][/tex]
Hence, based on the simplifications and results:
The integral results are:
[tex]\[ (a) \, \int \cos^3 \theta \, d\theta = -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
[tex]\[ (b) \, \int \frac{\ln \left(x^2\right)}{x^2} \, dx = -\frac{\log(x^2)}{x} - \frac{2}{x} + C \][/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.