Worksheet for Section 7.1 - Integration by Parts The Integration ... - Free Printable

Let's solve each of these integrals using integration by parts, which follows the formula:

$$
\int u \, dv = uv - \int v \, du
$$

We'll go through each problem step by step.

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(a) $\displaystyle \int_0^\pi x \cos x \, dx$

Step 1: Choose $u$ and $dv$

Let:
- $u = x$ → $du = dx$
- $dv = \cos x \, dx$ → $v = \sin x$

Apply integration by parts:

$$
\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C
$$

Now evaluate from $0$ to $\pi$:

$$
\left[ x \sin x + \cos x \right]_0^\pi = (\pi \cdot \sin \pi + \cos \pi) - (0 \cdot \sin 0 + \cos 0)
= (0 + (-1)) - (0 + 1) = -1 - 1 = -2
$$

✔ Answer: $\boxed{-2}$

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(b) $\displaystyle \int 6x^2 \sin(3x) \, dx$

Factor out constant: $6 \int x^2 \sin(3x) \, dx$

Use integration by parts. Let’s set:
- $u = x^2$ → $du = 2x \, dx$
- $dv = \sin(3x) \, dx$ → $v = -\frac{1}{3} \cos(3x)$

Then:
$$
\int x^2 \sin(3x) \, dx = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{3} \int x \cos(3x) \, dx
$$

Now compute $\int x \cos(3x) \, dx$ using parts again.

Let:
- $u = x$ → $du = dx$
- $dv = \cos(3x) \, dx$ → $v = \frac{1}{3} \sin(3x)$

So:
$$
\int x \cos(3x) \, dx = \frac{1}{3} x \sin(3x) - \frac{1}{3} \int \sin(3x) \, dx = \frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) + C
$$

Plug back:
$$
\int x^2 \sin(3x) \, dx = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{3} \left( \frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) \right)
= -\frac{1}{3} x^2 \cos(3x) + \frac{2}{9} x \sin(3x) + \frac{2}{27} \cos(3x) + C
$$

Multiply by 6:
$$
6 \int x^2 \sin(3x) \, dx = 6 \left( -\frac{1}{3} x^2 \cos(3x) + \frac{2}{9} x \sin(3x) + \frac{2}{27} \cos(3x) \right)
= -2x^2 \cos(3x) + \frac{4}{3} x \sin(3x) + \frac{4}{9} \cos(3x) + C
$$

✔ Answer: $\boxed{-2x^2 \cos(3x) + \frac{4}{3}x \sin(3x) + \frac{4}{9}\cos(3x) + C}$

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(c) $\displaystyle \int (x^2 - 4x)e^{2x} \, dx$

Let $u = x^2 - 4x$, $dv = e^{2x} dx$

Then:
- $du = (2x - 4) dx$
- $v = \frac{1}{2} e^{2x}$

So:
$$
\int (x^2 - 4x)e^{2x} dx = \frac{1}{2}(x^2 - 4x)e^{2x} - \frac{1}{2} \int (2x - 4)e^{2x} dx
$$

Now compute $\int (2x - 4)e^{2x} dx$

Let $u = 2x - 4$, $dv = e^{2x} dx$

- $du = 2 dx$
- $v = \frac{1}{2} e^{2x}$

$$
\int (2x - 4)e^{2x} dx = \frac{1}{2}(2x - 4)e^{2x} - \frac{1}{2} \int 2e^{2x} dx
= (x - 2)e^{2x} - \frac{1}{2} \cdot 2 \cdot \frac{1}{2} e^{2x}
= (x - 2)e^{2x} - \frac{1}{2} e^{2x}
$$

Now plug back:
$$
\int (x^2 - 4x)e^{2x} dx = \frac{1}{2}(x^2 - 4x)e^{2x} - \frac{1}{2} \left[(x - 2)e^{2x} - \frac{1}{2} e^{2x} \right]
= \frac{1}{2}(x^2 - 4x)e^{2x} - \frac{1}{2}(x - 2)e^{2x} + \frac{1}{4} e^{2x}
$$

Factor $e^{2x}$:
$$
= e^{2x} \left[ \frac{1}{2}(x^2 - 4x) - \frac{1}{2}(x - 2) + \frac{1}{4} \right]
= e^{2x} \left[ \frac{1}{2}x^2 - 2x - \frac{1}{2}x + 1 + \frac{1}{4} \right]
= e^{2x} \left( \frac{1}{2}x^2 - \frac{5}{2}x + \frac{5}{4} \right) + C
$$

✔ Answer: $\boxed{e^{2x} \left( \frac{1}{2}x^2 - \frac{5}{2}x + \frac{5}{4} \right) + C}$

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(d) $\displaystyle \int_1^{e^2} x \ln x \, dx$

Let:
- $u = \ln x$ → $du = \frac{1}{x} dx$
- $dv = x dx$ → $v = \frac{1}{2}x^2$

Then:
$$
\int x \ln x \, dx = \frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} dx
= \frac{1}{2}x^2 \ln x - \frac{1}{2} \int x \, dx
= \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + C
$$

Evaluate from $1$ to $e^2$:

At $x = e^2$:
$$
\frac{1}{2}(e^4)\ln(e^2) - \frac{1}{4}e^4 = \frac{1}{2}e^4 \cdot 2 - \frac{1}{4}e^4 = e^4 - \frac{1}{4}e^4 = \frac{3}{4}e^4
$$

At $x = 1$:
$$
\frac{1}{2}(1)^2 \ln(1) - \frac{1}{4}(1)^2 = 0 - \frac{1}{4} = -\frac{1}{4}
$$

Subtract:
$$
\frac{3}{4}e^4 - (-\frac{1}{4}) = \frac{3}{4}e^4 + \frac{1}{4}
$$

✔ Answer: $\boxed{\frac{3}{4}e^4 + \frac{1}{4}}$

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(e) $\displaystyle \int \arccos(2x) \, dx$

Let:
- $u = \arccos(2x)$ → $du = -\frac{2}{\sqrt{1 - (2x)^2}} dx = -\frac{2}{\sqrt{1 - 4x^2}} dx$
- $dv = dx$ → $v = x$

Then:
$$
\int \arccos(2x) dx = x \arccos(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}} dx
$$

Now compute $\int \frac{2x}{\sqrt{1 - 4x^2}} dx$

Let $w = 1 - 4x^2$, $dw = -8x dx$, so $x dx = -\frac{1}{8} dw$

$$
\int \frac{2x}{\sqrt{1 - 4x^2}} dx = 2 \int \frac{x dx}{\sqrt{1 - 4x^2}} = 2 \cdot \left(-\frac{1}{8}\right) \int w^{-1/2} dw = -\frac{1}{4} \cdot 2 w^{1/2} = -\frac{1}{2} \sqrt{1 - 4x^2}
$$

Wait: let's recheck:

$$
\int \frac{2x}{\sqrt{1 - 4x^2}} dx
$$

Let $u = 1 - 4x^2$, $du = -8x dx$, so $x dx = -\frac{1}{8} du$

Then:
$$
\int \frac{2x}{\sqrt{1 - 4x^2}} dx = 2 \int \frac{x dx}{\sqrt{1 - 4x^2}} = 2 \cdot \left( -\frac{1}{8} \int u^{-1/2} du \right)
= -\frac{1}{4} \cdot 2 u^{1/2} = -\frac{1}{2} \sqrt{1 - 4x^2}
$$

Yes.

So:
$$
\int \arccos(2x) dx = x \arccos(2x) - \frac{1}{2} \sqrt{1 - 4x^2} + C
$$

✔ Answer: $\boxed{x \arccos(2x) - \frac{1}{2} \sqrt{1 - 4x^2} + C}$

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(f) $\displaystyle \int_0^{1/2} \arcsin(2x) \, dx$

Let:
- $u = \arcsin(2x)$ → $du = \frac{2}{\sqrt{1 - 4x^2}} dx$
- $dv = dx$ → $v = x$

Then:
$$
\int \arcsin(2x) dx = x \arcsin(2x) - \int \frac{2x}{\sqrt{1 - 4x^2}} dx
$$

We already computed this integral in part (e):

$$
\int \frac{2x}{\sqrt{1 - 4x^2}} dx = -\frac{1}{2} \sqrt{1 - 4x^2}
$$

So:
$$
\int \arcsin(2x) dx = x \arcsin(2x) + \frac{1}{2} \sqrt{1 - 4x^2} + C
$$

Now evaluate from $0$ to $1/2$:

At $x = 1/2$:
- $\arcsin(2 \cdot 1/2) = \arcsin(1) = \frac{\pi}{2}$
- $x \arcsin(2x) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$
- $\sqrt{1 - 4(1/4)} = \sqrt{1 - 1} = 0$

So total: $\frac{\pi}{4} + \frac{1}{2} \cdot 0 = \frac{\pi}{4}$

At $x = 0$:
- $x \arcsin(0) = 0$
- $\sqrt{1 - 0} = 1$, so $\frac{1}{2} \cdot 1 = \frac{1}{2}$

So value: $0 + \frac{1}{2} = \frac{1}{2}$

Now subtract: $\frac{\pi}{4} - \frac{1}{2}$

✔ Answer: $\boxed{\frac{\pi}{4} - \frac{1}{2}}$

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(g) $\displaystyle \int \arctan(6x) \, dx$

Let:
- $u = \arctan(6x)$ → $du = \frac{6}{1 + (6x)^2} dx = \frac{6}{1 + 36x^2} dx$
- $dv = dx$ → $v = x$

Then:
$$
\int \arctan(6x) dx = x \arctan(6x) - \int \frac{6x}{1 + 36x^2} dx
$$

Now compute $\int \frac{6x}{1 + 36x^2} dx$

Let $u = 1 + 36x^2$, $du = 72x dx$, so $x dx = \frac{1}{72} du$

$$
\int \frac{6x}{1 + 36x^2} dx = 6 \cdot \frac{1}{72} \int \frac{1}{u} du = \frac{1}{12} \ln|u| = \frac{1}{12} \ln(1 + 36x^2)
$$

So:
$$
\int \arctan(6x) dx = x \arctan(6x) - \frac{1}{12} \ln(1 + 36x^2) + C
$$

✔ Answer: $\boxed{x \arctan(6x) - \frac{1}{12} \ln(1 + 36x^2) + C}$

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(h) $\displaystyle \int e^{2x} \sin(3x) \, dx$

This is a classic case requiring two applications of integration by parts.

Let:
- $u = \sin(3x)$ → $du = 3\cos(3x) dx$
- $dv = e^{2x} dx$ → $v = \frac{1}{2}e^{2x}$

Then:
$$
\int e^{2x} \sin(3x) dx = \frac{1}{2}e^{2x} \sin(3x) - \frac{3}{2} \int e^{2x} \cos(3x) dx
$$

Now compute $\int e^{2x} \cos(3x) dx$

Let:
- $u = \cos(3x)$ → $du = -3\sin(3x) dx$
- $dv = e^{2x} dx$ → $v = \frac{1}{2}e^{2x}$

So:
$$
\int e^{2x} \cos(3x) dx = \frac{1}{2}e^{2x} \cos(3x) + \frac{3}{2} \int e^{2x} \sin(3x) dx
$$

Now substitute back into earlier expression:

Let $I = \int e^{2x} \sin(3x) dx$

Then:
$$
I = \frac{1}{2}e^{2x} \sin(3x) - \frac{3}{2} \left[ \frac{1}{2}e^{2x} \cos(3x) + \frac{3}{2} I \right]
= \frac{1}{2}e^{2x} \sin(3x) - \frac{3}{4}e^{2x} \cos(3x) - \frac{9}{4} I
$$

Now solve for $I$:

$$
I + \frac{9}{4}I = \frac{1}{2}e^{2x} \sin(3x) - \frac{3}{4}e^{2x} \cos(3x)
\Rightarrow \frac{13}{4}I = e^{2x} \left( \frac{1}{2} \sin(3x) - \frac{3}{4} \cos(3x) \right)
$$

Multiply both sides by 4:
$$
13I = 4e^{2x} \left( \frac{1}{2} \sin(3x) - \frac{3}{4} \cos(3x) \right) = e^{2x} (2 \sin(3x) - 3 \cos(3x))
$$

So:
$$
I = \frac{e^{2x}}{13} (2 \sin(3x) - 3 \cos(3x)) + C
$$

✔ Answer: $\boxed{\frac{e^{2x}}{13} (2 \sin(3x) - 3 \cos(3x)) + C}$

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✔ Final Answers Summary:

(a) $\boxed{-2}$
(b) $\boxed{-2x^2 \cos(3x) + \frac{4}{3}x \sin(3x) + \frac{4}{9}\cos(3x) + C}$
(c) $\boxed{e^{2x} \left( \frac{1}{2}x^2 - \frac{5}{2}x + \frac{5}{4} \right) + C}$
(d) $\boxed{\frac{3}{4}e^4 + \frac{1}{4}}$
(e) $\boxed{x \arccos(2x) - \frac{1}{2} \sqrt{1 - 4x^2} + C}$
(f) $\boxed{\frac{\pi}{4} - \frac{1}{2}}$
(g) $\boxed{x \arctan(6x) - \frac{1}{12} \ln(1 + 36x^2) + C}$
(h) $\boxed{\frac{e^{2x}}{13} (2 \sin(3x) - 3 \cos(3x)) + C}$

Let me know if you'd like any visual explanation or alternative methods!