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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]
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