Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Sure, let's evaluate the given integrals step-by-step.
### (a) [tex]\(\int \cos^3 \theta \, d\theta\)[/tex]
To begin, we need to express [tex]\(\cos^3 \theta\)[/tex] in a form that is easier to integrate. We can use the trigonometric identity:
[tex]\[ \cos^3 \theta = \cos \theta \cdot \cos^2 \theta \][/tex]
and we can rewrite [tex]\(\cos^2 \theta\)[/tex] using the identity [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]:
[tex]\[ \cos^3 \theta = \cos \theta \cdot (1 - \sin^2 \theta) = \cos \theta - \cos \theta \cdot \sin^2 \theta \][/tex]
Now, we can break this into two separate integrals:
[tex]\[ \int \cos^3 \theta \, d\theta = \int (\cos \theta - \cos \theta \cdot \sin^2 \theta) \, d\theta \][/tex]
We can split the integral:
[tex]\[ \int \cos^3 \theta \, d\theta = \int \cos \theta \, d\theta - \int \cos \theta \cdot \sin^2 \theta \, d\theta \][/tex]
The first integral is straightforward:
[tex]\[ \int \cos \theta \, d\theta = \sin \theta \][/tex]
For the second integral, let's use the substitution [tex]\( u = \sin \theta \)[/tex]. Then [tex]\( du = \cos \theta \, d\theta \)[/tex]:
[tex]\[ \int \cos \theta \cdot \sin^2 \theta \, d\theta = \int \sin^2 \theta \, du = \int u^2 \, du \][/tex]
Integrating [tex]\( u^2 \)[/tex]:
[tex]\[ \int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3 \theta}{3} + C \][/tex]
Combining these results, we have:
[tex]\[ \int \cos^3 \theta \, d\theta = \sin \theta - \frac{\sin^3 \theta}{3} + C \][/tex]
So, the result is:
[tex]\[ \int \cos^3 \theta \, d\theta = -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
### (b) [tex]\(\int \frac{\ln \left(x^2\right)}{x^2} \, dx\)[/tex]
First, simplify the integrand using the property of logarithms:
[tex]\[ \ln(x^2) = 2 \ln(x) \][/tex]
So the integral becomes:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = \int \frac{2 \ln(x)}{x^2} \, dx = 2 \int \frac{\ln(x)}{x^2} \, dx \][/tex]
Let's use the substitution [tex]\( u = \ln(x) \)[/tex]. Then [tex]\( du = \frac{1}{x} \, dx \)[/tex], which implies [tex]\( dx = x \, du \)[/tex]:
[tex]\[ 2 \int \frac{\ln(x)}{x^2} \, dx = 2 \int \frac{u}{x} \cdot \frac{1}{x} \, dx = 2 \int \frac{u}{x^2} \cdot x \, du = 2 \int \frac{u}{x} \cdot x \, du \][/tex]
Since [tex]\( x \, du = dx \)[/tex], we have:
[tex]\[ 2 \int u \cdot \frac{du}{x} = 2 \int u \cdot \frac{du}{x} \][/tex]
Given [tex]\( x = e^u \)[/tex], we get [tex]\( \frac{du}{x} = e^{-u} \, du \)[/tex]:
[tex]\[ 2 \int u \cdot e^{-u} \, du \][/tex]
Integrate by parts, taking [tex]\( v = u \)[/tex] and [tex]\( dw = e^{-u} \, du \)[/tex]:
[tex]\[ dv = du \text{ and } w = -e^{-u} \][/tex]
Applying the integration by parts formula [tex]\( \int v \, dw = vw - \int w \, dv \)[/tex]:
[tex]\[ 2 \int u \cdot e^{-u} \, du = 2 \left( -u e^{-u} - \int -e^{-u} \, du \right) = 2 \left( -u e^{-u} + e^{-u} \right) = 2 \left( \frac{-\ln(x)}{x} + \frac{1}{x} \right) \][/tex]
So, the integral is:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = 2 \left( -\frac{\ln(x)}{x} + \frac{1}{x} \right) + C = \frac{-2 \ln(x)}{x} + \frac{2}{x} + C \][/tex]
Thus, the final answer is:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = -\frac{\ln(x^2)}{x} - \frac{2}{x} + C \][/tex]
### (a) [tex]\(\int \cos^3 \theta \, d\theta\)[/tex]
To begin, we need to express [tex]\(\cos^3 \theta\)[/tex] in a form that is easier to integrate. We can use the trigonometric identity:
[tex]\[ \cos^3 \theta = \cos \theta \cdot \cos^2 \theta \][/tex]
and we can rewrite [tex]\(\cos^2 \theta\)[/tex] using the identity [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]:
[tex]\[ \cos^3 \theta = \cos \theta \cdot (1 - \sin^2 \theta) = \cos \theta - \cos \theta \cdot \sin^2 \theta \][/tex]
Now, we can break this into two separate integrals:
[tex]\[ \int \cos^3 \theta \, d\theta = \int (\cos \theta - \cos \theta \cdot \sin^2 \theta) \, d\theta \][/tex]
We can split the integral:
[tex]\[ \int \cos^3 \theta \, d\theta = \int \cos \theta \, d\theta - \int \cos \theta \cdot \sin^2 \theta \, d\theta \][/tex]
The first integral is straightforward:
[tex]\[ \int \cos \theta \, d\theta = \sin \theta \][/tex]
For the second integral, let's use the substitution [tex]\( u = \sin \theta \)[/tex]. Then [tex]\( du = \cos \theta \, d\theta \)[/tex]:
[tex]\[ \int \cos \theta \cdot \sin^2 \theta \, d\theta = \int \sin^2 \theta \, du = \int u^2 \, du \][/tex]
Integrating [tex]\( u^2 \)[/tex]:
[tex]\[ \int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3 \theta}{3} + C \][/tex]
Combining these results, we have:
[tex]\[ \int \cos^3 \theta \, d\theta = \sin \theta - \frac{\sin^3 \theta}{3} + C \][/tex]
So, the result is:
[tex]\[ \int \cos^3 \theta \, d\theta = -\frac{\sin^3 \theta}{3} + \sin \theta + C \][/tex]
### (b) [tex]\(\int \frac{\ln \left(x^2\right)}{x^2} \, dx\)[/tex]
First, simplify the integrand using the property of logarithms:
[tex]\[ \ln(x^2) = 2 \ln(x) \][/tex]
So the integral becomes:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = \int \frac{2 \ln(x)}{x^2} \, dx = 2 \int \frac{\ln(x)}{x^2} \, dx \][/tex]
Let's use the substitution [tex]\( u = \ln(x) \)[/tex]. Then [tex]\( du = \frac{1}{x} \, dx \)[/tex], which implies [tex]\( dx = x \, du \)[/tex]:
[tex]\[ 2 \int \frac{\ln(x)}{x^2} \, dx = 2 \int \frac{u}{x} \cdot \frac{1}{x} \, dx = 2 \int \frac{u}{x^2} \cdot x \, du = 2 \int \frac{u}{x} \cdot x \, du \][/tex]
Since [tex]\( x \, du = dx \)[/tex], we have:
[tex]\[ 2 \int u \cdot \frac{du}{x} = 2 \int u \cdot \frac{du}{x} \][/tex]
Given [tex]\( x = e^u \)[/tex], we get [tex]\( \frac{du}{x} = e^{-u} \, du \)[/tex]:
[tex]\[ 2 \int u \cdot e^{-u} \, du \][/tex]
Integrate by parts, taking [tex]\( v = u \)[/tex] and [tex]\( dw = e^{-u} \, du \)[/tex]:
[tex]\[ dv = du \text{ and } w = -e^{-u} \][/tex]
Applying the integration by parts formula [tex]\( \int v \, dw = vw - \int w \, dv \)[/tex]:
[tex]\[ 2 \int u \cdot e^{-u} \, du = 2 \left( -u e^{-u} - \int -e^{-u} \, du \right) = 2 \left( -u e^{-u} + e^{-u} \right) = 2 \left( \frac{-\ln(x)}{x} + \frac{1}{x} \right) \][/tex]
So, the integral is:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = 2 \left( -\frac{\ln(x)}{x} + \frac{1}{x} \right) + C = \frac{-2 \ln(x)}{x} + \frac{2}{x} + C \][/tex]
Thus, the final answer is:
[tex]\[ \int \frac{\ln(x^2)}{x^2} \, dx = -\frac{\ln(x^2)}{x} - \frac{2}{x} + C \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
