IDNLearn.com: Your reliable source for finding precise answers. Get accurate and detailed answers to your questions from our knowledgeable and dedicated community members.
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]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.
