Evaluate the following: 1. ∫√7x−1 dx 2. ∫sin(4πx + 7) dx 3. ∫√sin x cos x dx 4. ∫x cos x dx 5. ∫e^x sin x dx

Question

ScanSolve AI Tutor · Accepted Answer

Here are the evaluations of the integrals:

For $\int \sqrt{7x-1}\,dx$ :

Let $u = 7x-1$ .
Then $du = 7\,dx$ , which means $dx = \frac{1}{7}\,du$ .

Substitute $u$ into the integral.

$$
\int \sqrt{u} \cdot \frac{1}{7}\,du = \frac{1}{7} \int u^{1/2}\,du
$$

Integrate $u^{1/2}$ .

$$
\frac{1}{7} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{7} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{21} u^{3/2} + C
$$

Substitute back $u = 7x-1$ .
$$ \boxed{\frac{2}{21} (7x-1)^{3/2} + C} $$
2. For $\int \sin(4\pi x + 7)\,dx$ :
Let $u = 4\pi x + 7$ .
Then $du = 4\pi\,dx$ , which means $dx = \frac{1}{4\pi}\,du$ .

Substitute $u$ into the integral.

$$
\int \sin(u) \cdot \frac{1}{4\pi}\,du = \frac{1}{4\pi} \int \sin(u)\,du
$$

Integrate $\sin(u)$ .

$$
\frac{1}{4\pi} (-\cos(u)) + C = -\frac{1}{4\pi} \cos(u) + C
$$

Substitute back $u = 4\pi x + 7$ .
$$ \boxed{-\frac{1}{4\pi} \cos(4\pi x + 7) + C} $$
3. For $\int \sqrt{\sin x} \cos x\,dx$ :
Let $u = \sin x$ .
Then $du = \cos x\,dx$ .
Substitute $u$ into the integral.
$$ \int \sqrt{u}\,du = \int u^{1/2}\,du $$

Integrate $u^{1/2}$ .

$$
\frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C
$$

Substitute back $u = \sin x$ .
$$ \boxed{\frac{2}{3} (\sin x)^{3/2} + C} $$
4. For $\int x \cos x\,dx$ :
Use integration by parts, $\int u\,dv = uv - \int v\,du$ .
Let $u = x$ and $dv = \cos x\,dx$ .
Find $du$ and $v$ .
$du = dx$ and $v = \int \cos x\,dx = \sin x$ .
Apply the integration by parts formula.
$$ \int x \cos x\,dx = x \sin x - \int \sin x\,dx $$
Integrate $\int \sin x\,dx$ .
$$ x \sin x - (-\cos x) + C = x \sin x + \cos x + C $$ $$ \boxed{x \sin x + \cos x + C} $$
5. For $\int e^x \sin x\,dx$ :
Use integration by parts, $\int u\,dv = uv - \int v\,du$ .
Let $I = \int e^x \sin x\,dx$ .
First application of integration by parts
Let $u = \sin x$ and $dv = e^x\,dx$ . Then $du = \cos x\,dx$ and $v = e^x$ . $I = e^x \sin x - \int e^x \cos x\,dx$
Second application of integration by parts for $\int e^x \cos x\,dx$
Let $u = \cos x$ and $dv = e^x\,dx$ . Then $du = -\sin x\,dx$ and $v = e^x$ . $\int e^x \cos x\,dx = e^x \cos x - \int e^x (-\sin x)\,dx = e^x \cos x + \int e^x \sin x\,dx$
Substitute this back into the equation for $I$ .
$$ I = e^x \sin x - (e^x \cos x + I) $$

Solve for $I$ .

$$
I = e^x \sin x - e^x \cos x - I
$$
$$
2I = e^x \sin x - e^x \cos x
$$
$$
I = \frac{1}{2} (e^x \sin x - e^x \cos x) + C
$$
$$
\boxed{\frac{e^x}{2} (\sin x - \cos x) + C}
$$

Evaluate the following: 1. ∫√7x−1 dx 2. ∫sin(4πx + 7) dx 3. ∫√sin x cos x dx 4. ∫x cos x dx 5. ∫e^x sin x dx

Step-by-step solution

Let $u = 7x-1$ .

Substitute $u$ into the integral.

Integrate $u^{1/2}$ .

Substitute back $u = 7x-1$ .

Let $u = 4\pi x + 7$ .

Substitute $u$ into the integral.

Integrate $\sin(u)$ .

Substitute back $u = 4\pi x + 7$ .

Let $u = \sin x$ .

Substitute $u$ into the integral.

Integrate $u^{1/2}$ .

Substitute back $u = \sin x$ .

Use integration by parts, $\int u\,dv = uv - \int v\,du$ .

Find $du$ and $v$ .

Apply the integration by parts formula.

Integrate $\int \sin x\,dx$ .

Use integration by parts, $\int u\,dv = uv - \int v\,du$ .

First application of integration by parts

Second application of integration by parts for $\int e^x \cos x\,dx$

Substitute this back into the equation for $I$ .

Solve for $I$ .

Still stuck on this one?

More Mathematics Questions

Evaluate the following: 1. ∫√7x−1 dx 2. ∫sin(4πx + 7) dx 3. ∫√sin x cos x dx 4. ∫x cos x dx 5. ∫e^x sin x dx

Step-by-step solution

Let u=7x−1u = 7x-1u=7x−1.

Substitute uuu into the integral.

Integrate u1/2u^{1/2}u1/2.

Substitute back u=7x−1u = 7x-1u=7x−1.

Let u=4πx+7u = 4\pi x + 7u=4πx+7.

Substitute uuu into the integral.

Integrate sin⁡(u)\sin(u)sin(u).

Substitute back u=4πx+7u = 4\pi x + 7u=4πx+7.

Let u=sin⁡xu = \sin xu=sinx.

Substitute uuu into the integral.

Integrate u1/2u^{1/2}u1/2.

Substitute back u=sin⁡xu = \sin xu=sinx.

Use integration by parts, ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du∫udv=uv−∫vdu.

Find dududu and vvv.

Apply the integration by parts formula.

Integrate ∫sin⁡x dx\int \sin x\,dx∫sinxdx.

Use integration by parts, ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du∫udv=uv−∫vdu.

First application of integration by parts

Second application of integration by parts for ∫excos⁡x dx\int e^x \cos x\,dx∫excosxdx

Substitute this back into the equation for III.

Solve for III.

Still stuck on this one?

More Mathematics Questions

Let $u = 7x-1$ .

Substitute $u$ into the integral.

Integrate $u^{1/2}$ .

Substitute back $u = 7x-1$ .

Let $u = 4\pi x + 7$ .

Substitute $u$ into the integral.

Integrate $\sin(u)$ .

Substitute back $u = 4\pi x + 7$ .

Let $u = \sin x$ .

Substitute $u$ into the integral.

Integrate $u^{1/2}$ .

Substitute back $u = \sin x$ .

Use integration by parts, $\int u\,dv = uv - \int v\,du$ .

Find $du$ and $v$ .

Integrate $\int \sin x\,dx$ .

Use integration by parts, $\int u\,dv = uv - \int v\,du$ .

Second application of integration by parts for $\int e^x \cos x\,dx$

Substitute this back into the equation for $I$ .

Solve for $I$ .