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INTEGRATION�BY �S.K. MOHANTY

MAYURBHANJ SCHOOL OF ENGINEERING

Semester-ll

Department- Sc & Hum

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TECHNIQUES OF INTEGRATION

Due to the Fundamental Theorem of Calculus (FTC), we can integrate a function if we know an antiderivative, that is, an indefinite integral.

We summarize the most important integrals �we have learned so far, as follows.

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FORMULAS OF INTEGRALS

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FORMULAS OF INTEGRALS

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FORMULAS OF INTEGRALS

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TECHNIQUES OF INTEGRATION

In this chapter, we develop techniques for using the basic integration formulas.

This helps obtain indefinite integrals of �more complicated functions.

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TECHNIQUES OF INTEGRATION

We learned the most important method �of integration, the Substitution Rule, in �Section 5.5

The other general technique, integration by �parts, is presented in Section 7.1

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TECHNIQUES OF INTEGRATION

Then, we learn methods that are special �to particular classes of functions—such as trigonometric functions and rational functions.

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TECHNIQUES OF INTEGRATION

Integration is not as straightforward �as differentiation.

There are no rules that absolutely guarantee �obtaining an indefinite integral of a function.

Therefore, we discuss a strategy for integration �in Section 7.5

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7.1�Integration by Parts

In this section, we will learn:

How to integrate complex functions by parts.

TECHNIQUES OF INTEGRATION

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INTEGRATION BY PARTS

Every differentiation rule has �a corresponding integration rule.

For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation.

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INTEGRATION BY PARTS

The rule that corresponds to �the Product Rule for differentiation �is called the rule for integration by �parts.

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INTEGRATION BY PARTS

The Product Rule states that, if f and g �are differentiable functions, then

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INTEGRATION BY PARTS

In the notation for indefinite integrals, �this equation becomes

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INTEGRATION BY PARTS

We can rearrange this equation as:

Formula 1

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INTEGRATION BY PARTS

Formula 1 is called the formula for �integration by parts.

It is perhaps easier to remember in �the following notation.

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INTEGRATION BY PARTS

Let u = f(x) and v = g(x).

Then, the differentials are: �� du = f’(x) dx and dv = g’(x) dx

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INTEGRATION BY PARTS

Thus, by the Substitution Rule, �the formula for integration by parts becomes:

Formula 2

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INTEGRATION BY PARTS

Find ∫ x sin x dx

Suppose we choose f(x) = x and g’(x) = sin x.

Then, f’(x) = 1 and g(x) = –cos x.

For g, we can choose any antiderivative of g’.

E. g. 1—Solution 1

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INTEGRATION BY PARTS

Using Formula 1, we have:

It’s wise to check the answer by differentiating it.
If we do so, we get x sin x, as expected.

E. g. 1—Solution 1

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INTEGRATION BY PARTS

Let
Then,

Using Formula 2, we have:

E. g. 1—Solution 2

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NOTE

Our aim in using integration by parts is �to obtain a simpler integral than the one �we started with.

Thus, in Example 1, we started with ∫ x sin x dx �and expressed it in terms of the simpler integral �∫ cos x dx.

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NOTE

If we had instead chosen u = sin x and �dv = x dx , then du = cos x dx and v = x²/2.
So, integration by parts gives:

Although this is true, ∫ x²cos x dx is a more difficult integral than the one we started with.

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NOTE

Hence, when choosing u and dv, we �usually try to keep u = f(x) to be a function �that becomes simpler when differentiated.

At least, it should not be more complicated.

However, make sure that dv = g’(x) dx can be �readily integrated to give v.

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INTEGRATION BY PARTS

Evaluate ∫ ln x dx

Here, we don’t have much choice for u and dv.

Then,

Example 2

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INTEGRATION BY PARTS

Integrating by parts, we get:

Example 2

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INTEGRATION BY PARTS

Integration by parts is effective in �this example because the derivative of �the function f(x) = ln x is simpler than f.

Example 2

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INTEGRATION BY PARTS

Find ∫ t²e^tdt

Notice that t²becomes simpler when differentiated.

However, e^tis unchanged when differentiated �or integrated.

Example 3

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INTEGRATION BY PARTS

So, we choose

Then,

Integration by parts gives:

E. g. 3—Equation 3

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INTEGRATION BY PARTS

The integral that we obtained, ∫ te^tdt, �is simpler than the original integral.
However, it is still not obvious.

So, we use integration by parts a second time.

Example 3

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INTEGRATION BY PARTS

This time, we choose � u = t and dv = e^tdt

Then, du = dt, v = e^t.

Example 3

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INTEGRATION BY PARTS

Putting this in Equation 3, we get

where C₁ = – 2C

Example 3

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INTEGRATION BY PARTS

Evaluate ∫ e^x sinx dx

e^x does not become simpler when differentiated.

Neither does sin x become simpler.

Example 4

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INTEGRATION BY PARTS

Nevertheless, we try choosing � u = e^x and dv = sin x

Then, du = e^x dx and v = – cos x.

E. g. 4—Equation 4

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INTEGRATION BY PARTS

So, integration by parts gives:

Example 4

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INTEGRATION BY PARTS

The integral we have obtained, ∫ e^xcos x dx, �is no simpler than the original one.

At least, it’s no more difficult.

Having had success in the preceding example integrating by parts twice, we do it again.

Example 4

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INTEGRATION BY PARTS

This time, we use � u = e^x and dv = cos x dx

Then, du = e^x dx, v = sin x, and

E. g. 4—Equation 5

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INTEGRATION BY PARTS

At first glance, it appears as if we have accomplished nothing.

We have arrived at ∫ e^x sin x dx, which is �where we started.

Example 4

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INTEGRATION BY PARTS

However, if we put the expression for �∫ e^x cos x dx from Equation 5 into Equation 4, we get:

This can be regarded as an equation to be �solved for the unknown integral.

Example 4

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INTEGRATION BY PARTS

Adding to both sides ∫ e^x sin x dx, �we obtain:

Example 4

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INTEGRATION BY PARTS

Dividing by 2 and adding the constant �of integration, we get:

Example 4

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INTEGRATION BY PARTS

The figure illustrates the example by �showing the graphs of f(x) = e^xsin x and �F(x) = ½ e^x(sin x – cos x).

As a visual check on �our work, notice that �f(x) = 0when F has a �maximum or minimum.

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INTEGRATION BY PARTS

If we combine the formula for integration �by parts with Part 2 of the FTC (FTC2), �we can evaluate definite integrals by parts.

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INTEGRATION BY PARTS

Evaluating both sides of Formula 1 between �a and b, assuming f’ and g’ are continuous, and using the FTC, we obtain:

Formula 6

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INTEGRATION BY PARTS

Calculate

Then,

Example 5

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INTEGRATION BY PARTS

So, Formula 6 gives:

Example 5

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INTEGRATION BY PARTS

To evaluate this integral, we use �the substitution t = 1 + x²(since u has another meaning in this example).

Then, dt = 2x dx.

So, x dx = ½ dt.

Example 5

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INTEGRATION BY PARTS

When x = 0, t = 1, and when x = 1, t = 2.

Hence,

Example 5

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INTEGRATION BY PARTS

Therefore,

Example 5

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INTEGRATION BY PARTS

As tan^-1x ≥ for x ≥ 0 , the integral in �the example can be interpreted as the area �of the region shown here.

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INTEGRATION BY PARTS

Prove the reduction formula

where n ≥ 2 is an integer.

This is called a reduction formula because �the exponent n has been reduced to n – 1 and n – 2.

E. g. 6—Formula 7

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INTEGRATION BY PARTS

Let
Then,

So, integration by parts gives:

Example 6

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INTEGRATION BY PARTS

Since cos²x = 1 – sin²x, we have:

As in Example 4, we solve this equation for �the desired integral by taking the last term on �the right side to the left side.

Example 6

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INTEGRATION BY PARTS

Thus, we have:

Example 6

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INTEGRATION BY PARTS

The reduction formula (7) is useful. �
By using it repeatedly, we could express �∫ sinⁿx dx in terms of:

∫ sin x dx (if n is odd)

∫ (sin x)⁰dx = ∫ dx (if n is even)