for each of the indefinite integrals below, select which of the following trig substitutions would be most helpful in evaluating the integral. do not evaluate the integrals.
a. ( x = 6 an heta )
b. ( x = 6sin heta )
c. ( x = 6sec heta )
1. ( int x^2sqrt{36 + x^2} dx )
2. ( int \frac{dx}{(36 - x^2)^{3/2}} )
3. ( int sqrt{x^2 - 36} dx )
4. ( int \frac{dx}{(36 + x^2)^3} )
5. ( int (x^2 - 36)^{5/2} dx )
note: in order to get credit for this problem all answers must be correct.
problem 3. (1 point)
which of the following ( u )-substitutions can be used to simplify the integral ( int an^7(x)sec^4(x) dx )?
select the best answer(s).
a. ( u = an(x)sec(x) )
b. ( u = sec^4(x) )
c. ( u = an^7(x) )
d. ( u = sec(x) )
e. ( u = an(x) )
f. ( u = sin(x) )

Question

for each of the indefinite integrals below, select which of the following trig substitutions would be most helpful in evaluating the integral. do not evaluate the integrals.
a. ( x = 6	an	heta )
b. ( x = 6sin	heta )
c. ( x = 6sec	heta )
1. ( int x^2sqrt{36 + x^2} dx )
2. ( int \frac{dx}{(36 - x^2)^{3/2}} )
3. ( int sqrt{x^2 - 36} dx )
4. ( int \frac{dx}{(36 + x^2)^3} )
5. ( int (x^2 - 36)^{5/2} dx )
note: in order to get credit for this problem all answers must be correct.
problem 3. (1 point)
which of the following ( u )-substitutions can be used to simplify the integral ( int 	an^7(x)sec^4(x) dx )?
select the best answer(s).
a. ( u = 	an(x)sec(x) )
b. ( u = sec^4(x) )
c. ( u = 	an^7(x) )
d. ( u = sec(x) )
e. ( u = 	an(x) )
f. ( u = sin(x) )

Accepted Answer

### Problem 1: $\int x^2\sqrt{36 + x^2}dx$
# Explanation:
## Step1: Identify the form
The integral has the form $\sqrt{a^2 + x^2}$ (here $a = 6$). The trigonometric substitution for $\sqrt{a^2 + x^2}$ is $x = a	an	heta$.
## Step2: Match with options
Given $a = 6$, the substitution is $x = 6	an	heta$, which is option A.
# Answer: A. $x = 6	an	heta$

### Problem 2: $\int \frac{dx}{(36 - x^2)^{3/2}}$
# Explanation:
## Step1: Identify the form
The integral has the form $\sqrt{a^2 - x^2}$ (here $a = 6$). The trigonometric substitution for $\sqrt{a^2 - x^2}$ is $x = a\sin	heta$.
## Step2: Match with options
Given $a = 6$, the substitution is $x = 6\sin	heta$, which is option B.
# Answer: B. $x = 6\sin	heta$

### Problem 3: $\int \sqrt{x^2 - 36}dx$
# Explanation:
## Step1: Identify the form
The integral has the form $\sqrt{x^2 - a^2}$ (here $a = 6$). The trigonometric substitution for $\sqrt{x^2 - a^2}$ is $x = a\sec	heta$.
## Step2: Match with options
Given $a = 6$, the substitution is $x = 6\sec	heta$, which is option C.
# Answer: C. $x = 6\sec	heta$

### Problem 4: $\int \frac{dx}{(36 + x^2)^3}$
# Explanation:
## Step1: Identify the form
The integral has the form $\sqrt{a^2 + x^2}$ (here $a = 6$) in the denominator (raised to a power). The trigonometric substitution for $\sqrt{a^2 + x^2}$ is $x = a	an	heta$.
## Step2: Match with options
Given $a = 6$, the substitution is $x = 6	an	heta$, which is option A.
# Answer: A. $x = 6	an	heta$

### Problem 5: $\int (x^2 - 36)^{5/2}dx$
# Explanation:
## Step1: Identify the form
The integral has the form $\sqrt{x^2 - a^2}$ (here $a = 6$) raised to a power. The trigonometric substitution for $\sqrt{x^2 - a^2}$ is $x = a\sec	heta$.
## Step2: Match with options
Given $a = 6$, the substitution is $x = 6\sec	heta$, which is option C.
# Answer: C. $x = 6\sec	heta$

### Problem 3 (the second Problem 3): $\int 	an^7(x)\sec^4(x)dx$
# Explanation:
## Step1: Rewrite the integrand
We can rewrite $\sec^4(x)$ as $\sec^2(x)\sec^2(x) = (1 + 	an^2(x))\sec^2(x)$. So the integrand becomes $	an^7(x)(1 + 	an^2(x))\sec^2(x)$.
## Step2: Analyze substitution
If we let $u = 	an(x)$, then $du = \sec^2(x)dx$. The integrand can be written in terms of $u$ as $u^7(1 + u^2)du$ (after substitution), which simplifies the integral. Also, if we let $u = \sec(x)$, $du=\sec(x)	an(x)dx$, but that doesn't match as well as $u = 	an(x)$. Let's check options:
- Option A: $u = 	an(x)\sec(x)$: $du = (\sec^2(x) + 	an^2(x)\sec(x))dx$, not helpful.
- Option B: $u = \sec^4(x)$: $du = 4\sec^4(x)	an(x)dx$, not matching the integrand.
- Option C: $u = 	an^7(x)$: $du = 7	an^6(x)\sec^2(x)dx$, not as simple as $u = 	an(x)$.
- Option D: $u = \sec(x)$: $du=\sec(x)	an(x)dx$, not matching the power of $	an(x)$.
- Option E: $u = 	an(x)$: $du = \sec^2(x)dx$, and as we rewrote the integrand, this substitution works.
- Option F: $u = \sin(x)$: $du = \cos(x)dx$, not related to $	an$ and $\sec$.

So the best substitution is $u = 	an(x)$ (option E) and also, since $\sec^4(x)=(1 + 	an^2(x))\sec^2(x)$, if we consider $u = \sec(x)$, but actually $u = 	an(x)$ is better. Wait, also, let's re - examine: $\sec^4(x)=\sec^2(x)\sec^2(x)=(1 + 	an^2(x))\sec^2(x)$. If we let $u=\sec(x)$, $du = \sec(x)	an(x)dx$, not helpful. If we let $u = 	an(x)$, then $\sec^2(x)=1 + u^2$, and $\sec^4(x)=(1 + u^2)^2$, and $du=\sec^2(x)dx$. So the integral becomes $\int u^7(1 + u^2)^2du$, which is easy to integrate. Also, if we consider the power of $\sec(x)$ is even, we can use $u = 	an(x)$. So options E (and also, let's check: if we let $u=\sec(x)$, no, but wait, the problem says "select the best answer(s)". Wait, actually, another way: $\sec^4(x)=\sec^2(x)\sec^2(x)$, and if we let $u = 	an(x)$, then $\sec^2(x)=1 + u^2$, so the integrand is $u^7(1 + u^2)\sec^2(x)dx$, and $du=\sec^2(x)dx$, so it becomes $\int u^7(1 + u^2)du$. Also, if we let $u=\sec(x)$, $du=\sec(x)	an(x)dx$, not matching. So the best substitution is $u = 	an(x)$ (option E) and also, wait, let's check the options again. Wait, the integrand is $	an^7(x)\sec^4(x)=	an^7(x)\sec^2(x)\sec^2(x)=	an^7(x)(1 + 	an^2(x))\sec^2(x)$. If we let $u = 	an(x)$, then $du=\sec^2(x)dx$, so the integral becomes $\int u^7(1 + u^2)du$, which is a polynomial in $u$. Also, if we let $u=\sec(x)$, $du=\sec(x)	an(x)dx$, which would require the integrand to have $\sec(x)	an(x)$ as a factor, but we have $	an^7(x)\sec^4(x)=	an^6(x)\sec^3(x)\cdot	an(x)\sec(x)$, so if we let $u=\sec(x)$, then $u^3	an^6(x)\cdot	an(x)\sec(x)dx$, and $	an^2(x)=u^2 - 1$, so $	an^6(x)=(u^2 - 1)^3$, so the integral becomes $\int (u^2 - 1)^3u^3\cdot udu=\int (u^2 - 1)^3u^4du$, which is also possible, but more complicated. So the best substitution is $u = 	an(x)$ (option E) and also, wait, the problem says "select the best answer(s)". Wait, maybe I made a mistake. Wait, the integrand is $	an^7(x)\sec^4(x)$. Let's write $\sec^4(x)$ as $\sec^2(x)\sec^2(x)=(1 + 	an^2(x))\sec^2(x)$. So if we let $u = 	an(x)$, then $du=\sec^2(x)dx$, and the integral becomes $\int u^7(1 + u^2)du$, which is a simple polynomial integral. So option E ($u = 	an(x)$) is correct. Also, is there another? Let's check option D: $u=\sec(x)$, $du=\sec(x)	an(x)dx$, then the integral is $\int 	an^7(x)\sec^4(x)dx=\int 	an^6(x)\sec^3(x)\cdot	an(x)\sec(x)dx=\int (u^2 - 1)^3u^3\cdot udu=\int (u^2 - 1)^3u^4du$, which is also possible, but more complex. So the best answer is E (and maybe D? Wait, no, let's check the options again. The problem says "select the best answer(s)". Wait, the original integrand: $	an^7(x)\sec^4(x)=	an^7(x)\sec^2(x)\sec^2(x)=	an^7(x)(1 + 	an^2(x))\sec^2(x)$. If we let $u = 	an(x)$, then $du=\sec^2(x)dx$, so the integral is $\int u^7(1 + u^2)du$, which is straightforward. So option E is correct. Also, if we let $u=\sec(x)$, it's more complicated. So the best answer is E.
# Answer: E. $u = 	an(x)$

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QUESTION IMAGE

Question

Explanation:

Problem 1: $\int x^2\sqrt{36 + x^2}dx$

Step1: Identify the form

Step2: Match with options

Step1: Identify the form

Step2: Match with options

Step1: Identify the form

Step2: Match with options

Answer:

Problem 2: $\int \frac{dx}{(36 - x^2)^{3/2}}$