problem 1. (1 point) for each of the following integrals, select the best ( u )-substitution to make. a : ( u = sin(x) ) b : ( u = cos(x) ) c : ( u = an(x) ) d : ( u = sec(x) ) 1. ( int 78 an^9(x) sec^{11}(x) dx ) 2. ( int 22 an^8(x) sec^{10}(x) dx ) note: you can earn partial credit on this problem.

Question

problem 1. (1 point) for each of the following integrals, select the best ( u )-substitution to make. a : ( u = sin(x) ) b : ( u = cos(x) ) c : ( u = 	an(x) ) d : ( u = sec(x) ) 1. ( int 78 	an^9(x) sec^{11}(x) dx ) 2. ( int 22 	an^8(x) sec^{10}(x) dx ) note: you can earn partial credit on this problem.

Accepted Answer

### Problem 1
# Explanation:
## Step1: Analyze the integral $\int 78	an^{9}(x)\sec^{11}(x)dx$
We know that the derivative of $\sec(x)$ is $\sec(x)	an(x)$, i.e., $du = \sec(x)	an(x)dx$ if $u=\sec(x)$. Let's rewrite the integral:
$\int 78	an^{9}(x)\sec^{11}(x)dx=78\int	an^{8}(x)\sec^{10}(x)\cdot\sec(x)	an(x)dx$
If we let $u = \sec(x)$, then $du=\sec(x)	an(x)dx$ and $	an^{2}(x)=\sec^{2}(x) - 1$, but here we can also note that the power of $\sec(x)$ is 11 and power of $	an(x)$ is 9. The derivative of $\sec(x)$ involves $\sec(x)	an(x)$, so when we substitute $u = \sec(x)$, the integral can be transformed. Also, if we consider the form of the integral which is a product of $	an^{n}(x)$ and $\sec^{m}(x)$, when $m$ is odd (here $m = 11$ is odd), we can split off a $\sec(x)	an(x)$ term for substitution. But also, if we let $u=\sec(x)$, then $du=\sec(x)	an(x)dx$, and we can rewrite $	an^{9}(x)=	an^{8}(x)	an(x)=(\sec^{2}(x)- 1)^{4}	an(x)$. But another way: Let's check the options. If we take $u = \sec(x)$, then $du=\sec(x)	an(x)dx$, and the integral becomes $78\int u^{10}\cdot	an^{8}(x)du$. But $	an^{8}(x)=(\sec^{2}(x)-1)^{4}=(u^{2}-1)^{4}$. However, also, if we consider the integral of the form $\int	an^{n}(x)\sec^{m}(x)dx$, when $m$ is even or odd, but in this case, the best substitution is $u = \sec(x)$? Wait, no, wait. Wait, the derivative of $	an(x)$ is $\sec^{2}(x)$, i.e., $du=\sec^{2}(x)dx$ if $u = 	an(x)$. Wait, let's re - examine.

Wait, the integral is $\int	an^{9}(x)\sec^{11}(x)dx$. Let's write $\sec^{11}(x)=\sec^{10}(x)\sec(x)$ and $	an^{9}(x)=	an^{8}(x)	an(x)$. Then $	an^{8}(x)=(\sec^{2}(x)-1)^{4}$. So if we let $u=\sec(x)$, then $du=\sec(x)	an(x)dx$, and $	an^{8}(x)=(\ u^{2}-1)^{4}$. Then the integral becomes $78\int(u^{2}-1)^{4}u^{10}du$, which is a polynomial in $u$. But if we let $u = 	an(x)$, then $du=\sec^{2}(x)dx$, and $\sec^{11}(x)=\sec^{9}(x)\sec^{2}(x)=(1 + 	an^{2}(x))^{4.5}\sec^{2}(x)$, which is not as nice. Wait, but the power of $\sec(x)$ is 11 (odd) and power of $	an(x)$ is 9 (odd). Wait, maybe I made a mistake. Let's recall the standard substitutions for $\int	an^{n}(x)\sec^{m}(x)dx$:

- If $m$ is even, we can let $u=	an(x)$ (since $du=\sec^{2}(x)dx$ and $\sec^{2}(x)=1+	an^{2}(x)$)
- If $n$ is odd, we can let $u = \sec(x)$ (since $du=\sec(x)	an(x)dx$ and we can split off a $\sec(x)	an(x)$ term)

In the first integral, $n = 9$ (odd) and $m = 11$ (odd). But let's check the options. The options are $u=\sin(x),\cos(x),	an(x),\sec(x)$.

Wait, let's compute the derivative of each option:

- If $u=\sin(x)$, $du=\cos(x)dx$, not related to the integrand.
- If $u=\cos(x)$, $du=-\sin(x)dx$, not related.
- If $u=	an(x)$, $du=\sec^{2}(x)dx$
- If $u=\sec(x)$, $du=\sec(x)	an(x)dx$

The integrand is $78	an^{9}(x)\sec^{11}(x)dx=78	an^{8}(x)\sec^{10}(x)\cdot\sec(x)	an(x)dx$. If we let $u = \sec(x)$, then $du=\sec(x)	an(x)dx$ and $	an^{8}(x)=(\sec^{2}(x)-1)^{4}=(u^{2}-1)^{4}$. So the integral becomes $78\int(u^{2}-1)^{4}u^{10}du$, which is a polynomial integral. Alternatively, if we let $u=	an(x)$, $du=\sec^{2}(x)dx$, and $\sec^{11}(x)=\sec^{9}(x)\sec^{2}(x)=(1 + 	an^{2}(x))^{4.5}\sec^{2}(x)$, which is more complicated. So the best substitution is $u=\sec(x)$? Wait, no, wait the power of $\sec(x)$ is 11 (odd) and power of $	an(x)$ is 9 (odd). Wait, maybe the answer is $u = \sec(x)$ (option D) for the first integral? Wait, no, let's check the second integral.

## Step2: Analyze the integral $\int22	an^{8}(x)\sec^{10}(x)dx$
$\int22	an^{8}(x)\sec^{10}(x)dx=22\int	an^{8}(x)\sec^{8}(x)\cdot\sec^{2}(x)dx$
If we let $u = 	an(x)$, then $du=\sec^{2}(x)dx$ and $\sec^{2}(x)=1 + 	an^{2}(x)=1+u^{2}$. So $\sec^{8}(x)=(1 + 	an^{2}(x))^{4}=(1 + u^{2})^{4}$. Then the integral becomes $22\int u^{8}(1 + u^{2})^{4}du$, which is a polynomial integral. Also, since the power of $\sec(x)$ is 10 (even), we can use the substitution $u=	an(x)$ (because $du=\sec^{2}(x)dx$ and we can express $\sec^{2k}(x)$ in terms of $	an(x)$).

Wait, let's go back to the first integral: $\int78	an^{9}(x)\sec^{11}(x)dx$. Let's write $	an^{9}(x)=	an^{8}(x)	an(x)=(\sec^{2}(x)-1)^{4}	an(x)$ and $\sec^{11}(x)=\sec^{10}(x)\sec(x)$. So the integral is $78\int(\sec^{2}(x)-1)^{4}\sec^{10}(x)\cdot\sec(x)	an(x)dx$. If we let $u=\sec(x)$, then $du=\sec(x)	an(x)dx$ and the integral becomes $78\int(u^{2}-1)^{4}u^{10}du$, which is a polynomial. But also, if we let $u=	an(x)$, $du=\sec^{2}(x)dx$, and $\sec^{11}(x)=\sec^{9}(x)\sec^{2}(x)=(1 + 	an^{2}(x))^{4.5}\sec^{2}(x)$, which is not a polynomial. So for the first integral, the best substitution is $u=\sec(x)$ (option D)? Wait, no, let's check the options again.

Wait, the first integral: $\int	an^{9}(x)\sec^{11}(x)dx$. Let's use $u = \sec(x)$, then $du=\sec(x)	an(x)dx$, so $	an^{9}(x)\sec^{11}(x)dx=	an^{8}(x)\sec^{10}(x)\cdot\sec(x)	an(x)dx=(\sec^{2}(x)-1)^{4}\sec^{10}(x)\cdot\sec(x)	an(x)dx=(u^{2}-1)^{4}u^{10}du$.

For the second integral: $\int	an^{8}(x)\sec^{10}(x)dx=\int	an^{8}(x)\sec^{8}(x)\sec^{2}(x)dx=\int	an^{8}(x)(1 + 	an^{2}(x))^{4}\sec^{2}(x)dx$. Let $u=	an(x)$, then $du=\sec^{2}(x)dx$, and the integral becomes $\int u^{8}(1 + u^{2})^{4}du$, which is a polynomial in $u$.

Wait, but let's check the options:

Option A: $u = \sin(x)$: derivative is $\cos(x)$, not related.

Option B: $u=\cos(x)$: derivative is $-\sin(x)$, not related.

Option C: $u=	an(x)$: derivative is $\sec^{2}(x)$

Option D: $u=\sec(x)$: derivative is $\sec(x)	an(x)$

For the first integral $\int78	an^{9}(x)\sec^{11}(x)dx$:

We can rewrite it as $78\int	an^{8}(x)\sec^{10}(x)\cdot\sec(x)	an(x)dx$. If we let $u = \sec(x)$, then $du=\sec(x)	an(x)dx$ and $	an^{8}(x)=(\sec^{2}(x)-1)^{4}=(u^{2}-1)^{4}$. So the integral becomes $78\int(u^{2}-1)^{4}u^{10}du$, which is a valid substitution.

For the second integral $\int22	an^{8}(x)\sec^{10}(x)dx$:

We can rewrite it as $22\int	an^{8}(x)\sec^{8}(x)\sec^{2}(x)dx=22\int	an^{8}(x)(1 + 	an^{2}(x))^{4}\sec^{2}(x)dx$. If we let $u=	an(x)$, then $du=\sec^{2}(x)dx$, and the integral becomes $22\int u^{8}(1 + u^{2})^{4}du$, which is a valid substitution. Wait, but the power of $\sec(x)$ is 10 (even), so using $u = 	an(x)$ is better because we can express $\sec^{2k}(x)$ in terms of $	an(x)$.

Wait, maybe I made a mistake in the first integral. Let's re - evaluate:

The integral $\int	an^{n}(x)\sec^{m}(x)dx$:

- If $m$ is even, use $u=	an(x)$ (since $du=\sec^{2}(x)dx$ and $\sec^{2}(x)=1 + 	an^{2}(x)$)
- If $n$ is odd, use $u=\sec(x)$ (since $du=\sec(x)	an(x)dx$ and we can split off a $\sec(x)	an(x)$ term)

In the first integral, $n = 9$ (odd) and $m = 11$ (odd). So we can use $u=\sec(x)$ (because $n$ is odd). In the second integral, $n = 8$ (even) and $m = 10$ (even). So we can use $u=	an(x)$ (because $m$ is even, we can factor out $\sec^{2}(x)$ for substitution with $u = 	an(x)$).

So for the first integral, the best substitution is $u=\sec(x)$ (option D), and for the second integral, the best substitution is $u=	an(x)$ (option C)? Wait, no, let's check the options again.

Wait, the first integral: $\int	an^{9}(x)\sec^{11}(x)dx$. Let's take $u=\sec(x)$, $du=\sec(x)	an(x)dx$. Then the integral becomes $\int	an^{8}(x)\sec^{10}(x)\cdot\sec(x)	an(x)dx=\int(\sec^{2}(x)-1)^{4}\sec^{10}(x)du=\int(u^{2}-1)^{4}u^{10}du$, which is a polynomial.

For the second integral: $\int	an^{8}(x)\sec^{10}(x)dx=\int	an^{8}(x)\sec^{8}(x)\sec^{2}(x)dx=\int	an^{8}(x)(1 + 	an^{2}(x))^{4}\sec^{2}(x)dx$. Let $u=	an(x)$, $du=\sec^{2}(x)dx$, then the integral becomes $\int u^{8}(1 + u^{2})^{4}du$, which is a polynomial.

So:

1. For $\int78	an^{9}(x)\sec^{11}(x)dx$, the best $u$-substitution is $u = \sec(x)$ (option D)
2. For $\int22	an^{8}(x)\sec^{10}(x)dx$, the best $u$-substitution is $u=	an(x)$ (option C)

# Answer:
1. D. $u = \sec(x)$
2. C. $u=	an(x)$

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Question

Explanation:

Problem 1

Step1: Analyze the integral $\int 78\tan^{9}(x)\sec^{11}(x)dx$

Step2: Analyze the integral $\int22\tan^{8}(x)\sec^{10}(x)dx$

Answer: