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Sagot :
Let's evaluate the two given integrals step-by-step.
### Part (a): [tex]\( \int \sqrt{4-x^2} \, dx \)[/tex]
To evaluate the integral [tex]\( \int \sqrt{4-x^2} \, dx \)[/tex], we can use a trigonometric substitution. Here, we notice that the integrand involves a square root of the form [tex]\( \sqrt{a^2 - x^2} \)[/tex] which suggests a trigonometric substitution.
1. Substitution:
Let [tex]\( x = 2 \sin \theta \)[/tex]. Then, [tex]\( dx = 2 \cos \theta \, d\theta \)[/tex].
2. Rewrite the Integral:
Substitute [tex]\( x = 2 \sin \theta \)[/tex] into the integral:
[tex]\[ \int \sqrt{4 - x^2} \, dx = \int \sqrt{4 - (2\sin \theta)^2} \cdot 2 \cos \theta \, d\theta = \int \sqrt{4 - 4\sin^2 \theta} \cdot 2 \cos \theta \, d\theta \][/tex]
Simplify the square root:
[tex]\[ \sqrt{4 - 4\sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta \][/tex]
Thus, the integral becomes:
[tex]\[ \int 2 \cos \theta \cdot 2 \cos \theta \, d\theta = \int 4 \cos^2 \theta \, d\theta \][/tex]
3. Simplify Using a Trigonometric Identity:
Use the double-angle identity: [tex]\( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \)[/tex]:
[tex]\[ \int 4 \cos^2 \theta \, d\theta = \int 4 \left( \frac{1 + \cos(2\theta)}{2} \right) d\theta = \int (2 + 2 \cos(2\theta)) \, d\theta \][/tex]
4. Integrate:
[tex]\[ \int 2 \, d\theta + \int 2 \cos(2\theta) \, d\theta \][/tex]
The integral of the first term is straightforward:
[tex]\[ \int 2 \, d\theta = 2\theta \][/tex]
For the second term, use the substitution [tex]\( u = 2\theta \)[/tex], [tex]\( du = 2 \, d\theta \)[/tex] i.e. [tex]\( d\theta = \frac{du}{2} \)[/tex]:
[tex]\[ \int 2 \cos(2\theta) \, d\theta = \int 2 \cos u \cdot \frac{du}{2} = \int \cos u \, du = \sin u \][/tex]
Substitute back [tex]\( u = 2\theta \)[/tex]:
[tex]\[ \sin(2\theta) \][/tex]
So, combining the results:
[tex]\[ \int \sqrt{4-x^2} \, dx = 2\theta + \sin(2\theta) + C \][/tex]
5. Substitute Back [tex]\( \theta \)[/tex]:
Recall [tex]\( x = 2 \sin \theta \)[/tex], thus [tex]\( \theta = \sin^{-1}\left(\frac{x}{2}\right) \)[/tex]:
[tex]\[ \sin(2\theta) = 2\sin \theta \cos \theta = 2 \left(\frac{x}{2}\right) \sqrt{1 - \left(\frac{x}{2}\right)^2} = x \sqrt{1 - \frac{x^2}{4}} = \frac{x \sqrt{4 - x^2}}{2} \][/tex]
Therefore:
[tex]\[ 2\theta = 2 \sin^{-1}\left(\frac{x}{2}\right) \][/tex]
Substitute these back:
[tex]\[ 2 \sin^{-1}\left(\frac{x}{2}\right) + \frac{x \sqrt{4 - x^2}}{2} + C \][/tex]
Thus, the evaluated integral in Part (a) is:
[tex]\[ \int \sqrt{4-x^2} \, dx = \frac{x\sqrt{4 - x^2}}{2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C \][/tex]
### Part (b): [tex]\( \int (1 + \tan x) \cos x \, dx \)[/tex]
To evaluate [tex]\( \int(1 + \tan x) \cos x \, dx \)[/tex]:
1. Distribute [tex]\( \cos x \)[/tex]:
[tex]\[ \int (1 + \tan x) \cos x \, dx = \int \cos x \, dx + \int \tan x \cos x \, dx \][/tex]
2. Evaluate Each Integral:
- The first integral is straightforward:
[tex]\[ \int \cos x \, dx = \sin x \][/tex]
- For the second integral, note that [tex]\( \tan x = \frac{\sin x}{\cos x} \)[/tex]:
[tex]\[ \int \tan x \cos x \, dx = \int \sin x \, dx = -\cos x \][/tex]
3. Combine the Results:
[tex]\[ \sin x - \cos x + C \][/tex]
Thus, the evaluated integral in Part (b) is:
[tex]\[ \int (1 + \tan x) \cos x \, dx = \sin x - \cos x + C \][/tex]
So the final answers are:
- (a): [tex]\(\int \sqrt{4-x^2} \, dx = \frac{x \sqrt{4-x^2}}{2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C\)[/tex]
- (b): [tex]\(\int (1+\tan x) \cos x \, dx = \sin x - \cos x + C\)[/tex]
### Part (a): [tex]\( \int \sqrt{4-x^2} \, dx \)[/tex]
To evaluate the integral [tex]\( \int \sqrt{4-x^2} \, dx \)[/tex], we can use a trigonometric substitution. Here, we notice that the integrand involves a square root of the form [tex]\( \sqrt{a^2 - x^2} \)[/tex] which suggests a trigonometric substitution.
1. Substitution:
Let [tex]\( x = 2 \sin \theta \)[/tex]. Then, [tex]\( dx = 2 \cos \theta \, d\theta \)[/tex].
2. Rewrite the Integral:
Substitute [tex]\( x = 2 \sin \theta \)[/tex] into the integral:
[tex]\[ \int \sqrt{4 - x^2} \, dx = \int \sqrt{4 - (2\sin \theta)^2} \cdot 2 \cos \theta \, d\theta = \int \sqrt{4 - 4\sin^2 \theta} \cdot 2 \cos \theta \, d\theta \][/tex]
Simplify the square root:
[tex]\[ \sqrt{4 - 4\sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta \][/tex]
Thus, the integral becomes:
[tex]\[ \int 2 \cos \theta \cdot 2 \cos \theta \, d\theta = \int 4 \cos^2 \theta \, d\theta \][/tex]
3. Simplify Using a Trigonometric Identity:
Use the double-angle identity: [tex]\( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \)[/tex]:
[tex]\[ \int 4 \cos^2 \theta \, d\theta = \int 4 \left( \frac{1 + \cos(2\theta)}{2} \right) d\theta = \int (2 + 2 \cos(2\theta)) \, d\theta \][/tex]
4. Integrate:
[tex]\[ \int 2 \, d\theta + \int 2 \cos(2\theta) \, d\theta \][/tex]
The integral of the first term is straightforward:
[tex]\[ \int 2 \, d\theta = 2\theta \][/tex]
For the second term, use the substitution [tex]\( u = 2\theta \)[/tex], [tex]\( du = 2 \, d\theta \)[/tex] i.e. [tex]\( d\theta = \frac{du}{2} \)[/tex]:
[tex]\[ \int 2 \cos(2\theta) \, d\theta = \int 2 \cos u \cdot \frac{du}{2} = \int \cos u \, du = \sin u \][/tex]
Substitute back [tex]\( u = 2\theta \)[/tex]:
[tex]\[ \sin(2\theta) \][/tex]
So, combining the results:
[tex]\[ \int \sqrt{4-x^2} \, dx = 2\theta + \sin(2\theta) + C \][/tex]
5. Substitute Back [tex]\( \theta \)[/tex]:
Recall [tex]\( x = 2 \sin \theta \)[/tex], thus [tex]\( \theta = \sin^{-1}\left(\frac{x}{2}\right) \)[/tex]:
[tex]\[ \sin(2\theta) = 2\sin \theta \cos \theta = 2 \left(\frac{x}{2}\right) \sqrt{1 - \left(\frac{x}{2}\right)^2} = x \sqrt{1 - \frac{x^2}{4}} = \frac{x \sqrt{4 - x^2}}{2} \][/tex]
Therefore:
[tex]\[ 2\theta = 2 \sin^{-1}\left(\frac{x}{2}\right) \][/tex]
Substitute these back:
[tex]\[ 2 \sin^{-1}\left(\frac{x}{2}\right) + \frac{x \sqrt{4 - x^2}}{2} + C \][/tex]
Thus, the evaluated integral in Part (a) is:
[tex]\[ \int \sqrt{4-x^2} \, dx = \frac{x\sqrt{4 - x^2}}{2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C \][/tex]
### Part (b): [tex]\( \int (1 + \tan x) \cos x \, dx \)[/tex]
To evaluate [tex]\( \int(1 + \tan x) \cos x \, dx \)[/tex]:
1. Distribute [tex]\( \cos x \)[/tex]:
[tex]\[ \int (1 + \tan x) \cos x \, dx = \int \cos x \, dx + \int \tan x \cos x \, dx \][/tex]
2. Evaluate Each Integral:
- The first integral is straightforward:
[tex]\[ \int \cos x \, dx = \sin x \][/tex]
- For the second integral, note that [tex]\( \tan x = \frac{\sin x}{\cos x} \)[/tex]:
[tex]\[ \int \tan x \cos x \, dx = \int \sin x \, dx = -\cos x \][/tex]
3. Combine the Results:
[tex]\[ \sin x - \cos x + C \][/tex]
Thus, the evaluated integral in Part (b) is:
[tex]\[ \int (1 + \tan x) \cos x \, dx = \sin x - \cos x + C \][/tex]
So the final answers are:
- (a): [tex]\(\int \sqrt{4-x^2} \, dx = \frac{x \sqrt{4-x^2}}{2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C\)[/tex]
- (b): [tex]\(\int (1+\tan x) \cos x \, dx = \sin x - \cos x + C\)[/tex]
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