Trigonometric Integrals 2

Now that we have learned strategies for solving integrals with factors of sine and cosine we can use similar techniques to solve integrals with factors of tangent and secant. Using the identity sec²x = 1 + tan²x we are able to convert even powers of secant to tangent and vice versa. Now we will consider two examples to illustrate two common strategies used to solve integrals of the form

Suppose we have an integral such as

Observing that (d/dx)tanx=sec²x we can separate a factor of sec²x and still be left with an even power of secant. Using the identity sec²x = 1 + tan²x we can convert the remaining sec²x to an expression involving tangent. Thus we have:

Then substitute u=tanx to obtain:

Note: Suppose we tried to use the substitution u=secx, then du=secxtanxdx. When we separate out a factor of secxtanx we are left with an odd power of tangent which is not easily converted to secant.

Consider the integral

Since (d/dx)secx=secxtanx we can separate a factor of secxtanx and still be left with an even power of tangent which we can easily convert to an expression involving secant using the identity sec²x = 1 + tan²x. Thus we have:

Then substitute u=secx to obtain:

Note: Suppose we tried to use the substitution u=tanx, then du=sec²xdx. When we separate out a factor of sec²x we are left with an odd power of secant which is not easily converted to tangent.

Strategy for Evaluating

(a)

If the power of secant is even (n=2k, k>2) save a factor of sec²x and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of tanx.

then substitute u=tanx.

(b)

If the power of tangent is odd (m=2k+1), save a factor of secxtanx and use the identity sec²x = 1 + tan²x to express the remaining factors in terms of secx.

then substitute u=secx.

Note: If the power of secant is even and the power of tangent is odd then either method will suffice, although there may be less work involved to use method (a) if the power of secant is smaller, and method (b) if the power of tangent is smaller.

Find the indefinite trigonometric integral

Solution:

Find the definite trigonometric integral

Solution:

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Trigonometric Integrals 1

Suppose we have an integral such as

The easy mistake is to simply make the substitution u=sinx, but then du=cosxdx. So in order to integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers of cosine we need an extra sinx factor. Thus for this example knowing we need an extra sinx factor to integrate powers of cosine we can separate one sine factor and convert the remaining sin⁴x to an expression involving cosine using the identity sin²x + cos²x = 1.

Now by using our knowledge of substitution we can evaluate the integral by letting u=cosx, then du=-sinxdx and

Now consider the integral

If we were to use the method from the previous example and separate one cosine factor we would be left with a factor of cosine of odd degree which isn't easily converted to sine. We must now consider the half angle formulas

Using the half angle formula for cos²x, we have:

Strategy for Evaluating

(a)

If the power of sine is odd (m=2k+1), save one sine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of cosine.

then substitute u=cosx.

(b)

If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin²x + cos²x = 1 to convert the remaining factors in terms of sine.

then substitute u=sinx.

(c)

If the powers of both sine and cosine are even then use the half angle identities.

In some cases it may be helpful to use the identity

Find the indefinite trigonometric integral

Solution:

Using the half angle formulas solve the indefinite trigonometric integral

Solution:

Find the definite trigonometric integral

Solution:

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Integration By Parts

Suppose that we have an integral such as

Similar to integrals solved using the substitution method, there are no general equations for this indefinite integral. However there do not appear to be any clear substitutions that could be made to simplify this integral. This brings us to an integration technique known as integration by parts, which will call upon our knowledge of the Product Rule for differentiation.

The Product Rule states: If f and g are differentiable functions, then

By taking the indefinite integral of both sides of the equation we have:

and we can rearrange this equation as

To make it easier to remember it is commonly written in the following notation. Let u=f(x) and v=g(x). Then the differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes:

From our previous example, if we let u=x and dv=cosx, then du=dx and v=sinx. If we substitute these values into the formula we have:

Note: By choosing u=x we obtain a simpler integral than we started with. Had we chose u=cosx and dv=x then du=-sinx and v=(1/2)x² so integration by parts gives:

This equation is correct, but the integral is more difficult than the one we started with.

When choosing u and dv always try to choose u=f(x) to be a function that becomes simpler when differentiated (or at least not more complicated) and to choose dv=g'(x) to be a function that can be easily integrated to give v.

Examples

Find the general indefinite integral by integration by parts 

Solution:

Evaluate the definite integral by integration by parts

Solution:

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Integrals of Symmetric Functions

If f(x) is continuous on [-a, a] and f is an even function, then

If f(x) is continuous on [-a, a] and f is an odd function, then

These properties of integrals of symmetric functions are very helpful when solving integration problems. Some of the more challenging problems can be solved quite simply by using this property.

Examples

Evaluate the definite integral of the symmetric function 

Solution:

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The Substitution Rule

Suppose that we have an integral such as

With our current knowledge of integration, we can't find the general equation of this indefinite integral. There are no antidifferentiation formulas for this type of integral. However, from our knowledge of differentiation, specifically the chain rule, we know that 4x³ is the derivative of the function within the square root, x⁴ + 7. We must also account for the chain rule when we are performing integration. To do this, we use the substitution rule.

The Substitution Rule states: if u = g(x) is a differentiable function and f is continuous on the range of g, then

Note: Recall that if u = g(x), then du = g'(x)dx. If we substitute u into the left side of the equation for g(x) and du for g'(x)dx, then we get the integral on the right side of the equation.

From our previous example, if we let u = (x⁴+7), then du = 4x³dx. If we substitutite these values into the integral, we get an integral that can be solved using the antidifferentiation formulas.

However, this answer is still in terms of u. We must substitute u = (x⁴+7) into the resulting function, so that it is a function of x, rather than u.

The substitution rule also applies to definite integrals. The Substitution Rule for Definite Integrals states: If f is continuous on the range of u = g(x) and g'(x) is continuous on [a,b], then

Examples

Find the general indefinite integrals using the substitution rule 

Solution:

Evaluate the definite integral using the substitution rule 

Solution:

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