AOS4 Topic 8: Integration by Parts

How to make the choice of which to integrate and which to differentiate?

Tip: Most of the time (unless other function cannot be integrated), choose the function which can be differentiated to obtain 0.

In this case, since we know how to differentiate and integrate both functions:

Let \( u = x \) and \( v' = \cos(x) \)

Therefore, \( u' = 1 \) and \( v = \sin(x) \)
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Now apply the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du \]

Thus, we obtain:

\[ \int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx = x \sin(x) + \cos(x) + C \]

where \( C \in \mathbb{R} \)
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Since \( x^2 \) can be easily differentiated to obtain 0 and we know how to integrate \( \sin(x) \)

Let \( u = x^2 \) and \( v' = \sin(x) \)

Therefore, \( u' = 2x \) and \( v = -\cos(x) \)

Applying integration by parts we obtain:

\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2\int x\cos(x) \, dx \]

As you can see, we do not know how to evaluate the resultant integral so we would need to do another by-parts calculation. However, we already worked this out from the previous example so we can directly substitute it in.
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Substituting our answer from the previous example into the above results in:

\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2(x\sin(x)+\cos(x)) + C, \, C \in \mathbb{R} \]

Then expand the brackets to obtain your final answer.

\[ \int x^2 \sin(x) \, dx = (2-x^2)\cos(x) + 2x\sin(x) + C, \, C \in \mathbb{R} \]
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Here we are forced to differentiate ln(x) and integrate x since we do not know how to integrate ln(x).

Therefore \( u = \ln(x) \) and \( v' = x \).

From this we obtain,

\( u' = \frac{1}{x} \) and \( v = \frac{1}{2}x^2 \)
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Therefore,

\[ \int x \ln(x) \, dx = \frac{1}{2}x^2 \ln(x) - \int \frac{x}{2} \cdot \frac{1}{2}x^2 \, dx \]

\[ \int x \ln(x) \, dx = \frac{1}{2}x^2 \ln(x) - \int \frac{1}{4}x^3 \, dx \]

\[ \int x \ln(x) \, dx = \frac{1}{2}x^2 \ln(x) - \frac{1}{4} \cdot \frac{x^4}{4} + C, \quad C \in \mathbb{R} \]
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Although it seems that there is only one function, you can think of it as a product of two functions, \( \arcsin(x) \) and 1.

Like in example 3, we have no idea how to integrate \( \arcsin(x) \), so we must differentiate \( \arcsin(x) \) and integrate 1.

Let \( u = \arcsin(x) \) and \( v' = 1 \)

Therefore, \( u' = \frac{1}{\sqrt{1-x^2}} \) and \( v = x \)
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Now applying the integration by parts formula we obtain:

\( \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \int \sqrt{1 - x^2} \, dx \)

Applying a u-substitution to help evaluate the 2nd integral (let \( u = x^2 \)) results in:

∴ \( \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \frac{1}{2} \int \frac{1}{1 - u} \, du \)

Finally, simplifying the above we obtain:

∴ \( \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \frac{1}{2}(1 - x^2) + C \), \( C \in \mathbb{R} \)
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Although we were able to make at least one term into a zero, or something we knew how to differentiate in the previous examples, no matter how many times we apply the integration by parts formula we will never obtain an integral we know how to integrate or a zero :(

As a result, we need to use another method to integrate this.

We will begin as normal. Let \( u = \sin(2x) \) and \( v' = \cos(3x) \) (It does not matter which you choose since you can integrate and differentiate both)

Therefore, \( u' = 2\cos(2x) \) and \( v = \frac{1}{3}\sin(3x) \)

Now apply the integration by parts formula:

\[ \begin{align*} \int \sin(2x)\cos(3x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) - 2\int \cos(2x)\cos(3x) \, dx \\ & = \frac{1}{3}\sin(2x)\sin(3x) - \frac{2}{3}\int \cos(2x)\sin(3x) \, dx \end{align*} \]

You may notice a pattern emerging where if we applied by parts to a \( \sin(mx)\cos(nx) \) expression we obtain the same thing but reversed.

Using this knowledge, let us try and re-create the original integral by applying this again.
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Now focusing on: \( \int \cos(2x)\sin(3x) \, dx \)

Let \( u = \cos(2x) \) and \( v' = \sin(3x) \)

Then, \( u' = -2\sin(2x) \) and \( v = -\frac{1}{3}\cos(3x) \)

Now applying the integration by parts formula we obtain:

\[ \begin{align*} \int \cos(2x)\sin(3x) \, dx & = -\frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \\ & = -\frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \end{align*} \]

Now substituting our result into our previous equation we obtain:

\[ \begin{align*} \int \sin(2x)\cos(3x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) - \frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \\ & - \frac{1}{3}\cos(2x)\cos(3x) - \frac{2}{3}\int \sin(2x)\cos(3x) \, dx \end{align*} \]

After expanding the brackets of the above equation, we obtain:

\[ \begin{align*} \int \sin(2x)\cos(2x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) + \frac{2}{9}\cos(2x)\cos(3x) + \frac{4}{9}\int \sin(2x)\cos(3x) \, dx \end{align*} \]

You may notice the same integral on either side, we can now treat it like a variable and re-arrange for it.

\[ \begin{align*} \frac{5}{9}\int \sin(2x)\cos(2x) \, dx & = \frac{1}{3}\sin(2x)\sin(3x) + \frac{2}{9}\cos(2x)\cos(3x) \end{align*} \]

Finally, we can evaluate our integral by simplifying the above expression.

Therefore, \[ \begin{align*} \int \sin(2x)\cos(2x) \, dx & = \frac{3}{5}\sin(2x)\sin(3x) + \frac{2}{5}\cos(2x)\cos(3x) + C, \quad C \in \mathbb{R} \end{align*} \]
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Solution:

We have to make a choice and let one of the functions in the product equal \( u \) and one equal \( \frac{dv}{dx} \).

As a general rule, we let \( u \) be the function which will become simpler when we differentiate it. In this case, it makes sense to let \( u = x^2 \) and \( \frac{dv}{dx} = e^{3x} \).

Then, \( \frac{du}{dx} = 2x \) and \( v = \int e^{3x} \, dx = \frac{1}{3} e^{3x} \).

Then, using the formula for integration by parts, \[ \int x^2 e^{3x} \, dx = \frac{1}{3} e^{3x} \cdot x^2 - \int \frac{1}{3} e^{3x} \cdot 2x \, dx \] \[ = \frac{1}{3} x^2 e^{3x} - \int \frac{2}{3} xe^{3x} \, dx \]

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The resulting integral is still a product. It is a product of the functions \( \frac{2}{3}x \) and \( e^{3x} \). We can use the formula again. This time we choose \( u = \frac{2}{3}x \) and \( \frac{dv}{dx} = e^{3x} \).

Then \( \frac{du}{dx} = \frac{2}{3} \) and \( v = \int e^{3x} \, dx = \frac{1}{3} e^{3x} \).

So, \[ \int x^2 e^{3x} \, dx = \frac{1}{3} x^2 e^{3x} - \left( \frac{2}{9} x e^{3x} - \frac{2}{27} e^{3x} \right) + C \]

where \( C \) is the constant of integration. So we have done integration by parts twice to arrive at our final answer.
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Solution:

Let \( u = e^x \) and \( dv/dx = \cos(x) \). Then \( du/dx = e^x \) and \( v = \sin(x) \).

So, using integration by parts, we obtain: \[ \int e^x \cos(x) \, dx = e^x \sin(x) - \int e^x \sin(x) \, dx \tag{1} \]

Similarly, we can use integration by parts to obtain: \[ \int e^x \sin(x) \, dx = -e^x \cos(x) + \int e^x \cos(x) \, dx \tag{2} \]
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Substitute (2) into (1) and then rearrange: \[ \int e^x \cos(x) \, dx = e^x \sin(x) - (-e^x \cos(x) + \int e^x \cos(x) \, dx) \] \[ 2\int e^x \cos(x) \, dx = e^x \sin(x) + e^x \cos(x) + c \]

Now dividing by 2 and renaming the constant gives: \[ \int e^x \cos(x) \, dx = \frac{1}{2} e^x (\sin(x) + \cos(x)) + c \]
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Using the formula

\[ \int x^3 e^x dx = I_3 \quad \text{(by the definition of } I_3 \text{)} \] \[ = x^3 e^x - 3I_2 \quad \text{(expressing } I_3 \text{ in terms of } I_2 \text{)} \] \[ = x^3 e^x - 3x^2 e^x - 2I_1 \quad \text{(expressing } I_2 \text{ in terms of } I_1 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6I_1 \]
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\[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6I_0 \quad \text{(expressing } I_1 \text{ in terms of } I_0 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6 \int e^x dx \quad \text{(by the definition of } I_0 \text{)} \] \[ = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + c \] \[ = (x^3 - 3x^2 + 6x - 6)e^x + c \]
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Solution: Let \( u = \log_e x \) and \( \frac{dv}{dx} = 1 \). Then \( \frac{du}{dx} = \frac{1}{x} \) and \( v = x \).

We have \[ \int_{1}^{2} \log_e x \, dx = x \log_e x \Bigg|_{1}^{2} - \int_{1}^{2} 1 \, dx \] \[ = x \log_e x \Bigg|_{1}^{2} - \frac{x}{2} \Bigg|_{1}^{2} \] \[ = 2\log_e 2 - (2 - 1) \] \[ = 2\log_e 2 - 1 \]

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$$ \int \sin(x^4)(4x - 3) \, dx $$

Integrate by parts: \[ \int f \, g' = f \, g - \int f' \, g \] \[ f = 4x - 3, \quad g' = \sin(x^4) \] \[ f' = 4, \quad g = -4\cos(x^4) \] \[ \begin{align*} &= -4\cos(x^4)(4x - 3) - \int -16\cos(x^4) \, dx \end{align*} \]

Now solving: \[ \int -16 \cos\left(\frac{x}{4}\right) \, dx \]

Substitute \( u = \frac{x}{4} \) \[ \implies du = \frac{1}{4} \, dx \]

Steps: \[ = -64 \int \cos(u) \, du \]

Now solving: \[ \int \cos(u) \, du \]

This is a standard integral: \[ = \sin(u) \]
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Plug in solved integrals: \[ -64 \int \cos(u) \, du = -64 \sin(u) \]

Undo substitution \( u = \frac{x}{4} \): \[ = -64 \sin\left(\frac{x}{4}\right) \]

Plug in solved integrals: \[ -4 \cos\left(\frac{x}{4}\right) \left(4x - 3\right) - \int -16 \cos\left(\frac{x}{4}\right) \, dx \]

The problem is solved: \[ \int \sin\left(\frac{x}{4}\right) \left(4x - 3\right) \, dx = 64 \sin\left(\frac{x}{4}\right) - 4 \cos\left(\frac{x}{4}\right) \left(4x - 3\right) + C \]

\[ \pi \int_{0}^{\pi} f(x) \, dx = -16\pi - 76\sqrt{2} - 12 \]= 6.197051864910682
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Proof

Let \( I_n = \int \cos^n x \, dx \).

a) Show that \( I_n = \frac{1}{n} \sin x \cos^{n-1} x + \frac{n-1}{n}I_{n-2} \).

Using integration by parts with \( u = \cos^{n-1} x \) and \( dv = \cos x \, dx \), we have:

\[ du = -(n-1)\cos^{n-2} x \sin x \, dx \]

\[ v = \sin x \]

Using the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]

We get: \[ I_n = \sin x \cos^{n-1} x - \int \sin x (- (n-1)\cos^{n-2} x \sin x) \, dx \] \[ I_n = \sin x \cos^{n-1} x + (n-1) \int \sin^2 x \cos^{n-2} x \, dx \]
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Now, using the identity \( \sin^2 x = 1 - \cos^2 x \), we rewrite the integral: \[ I_n = \sin x \cos^{n-1} x + (n-1) \int (1 - \cos^2 x) \cos^{n-2} x \, dx \] \[ I_n = \sin x \cos^{n-1} x + (n-1) \int \cos^{n-2} x \, dx - (n-1) \int \cos^n x \, dx \]

Notice that the second term \( \int \cos^{n-2} x \, dx \) is \( I_{n-2} \). So, we have: \[ I_n = \sin x \cos^{n-1} x + (n-1)I_{n-2} - (n-1)I_n \]

Solving for \( I_n \), we get: \[ I_n + (n-1)I_n = \sin x \cos^{n-1} x + (n-1)I_{n-2} \] \[ I_n(1 + n-1) = \sin x \cos^{n-1} x + (n-1)I_{n-2} \] \[ I_n = \frac{1}{n} \sin x \cos^{n-1} x + \frac{n-1}{n}I_{n-2} \]

This is the desired result.
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