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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]
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