Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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 hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
