Trigonometric Integrals: Techniques for Integrating Trigonometric Functions

Trigonometric Integrals (Part I)

In this section, let’s discuss the integration of trigonometric functions, and focus on two types of such integrals: those involving the sine and cosine functions in this form.

∫ sin ax sin bx dx

∫ sin ax cos bx dx

∫ cos ax cos bx dx

For example, the product of two sine functions with different multiples or one sine function and one cosine function with multiples can be represented as a trigonometric integral. The second type of trigonometric integral is the combination of sin x to a certain power together with cosine x to another power.

∫ sinmx cosnx dx, m, n≥0

Trigonometric integrals of the first type are easy to handle because we have compound angle formulas for the sine and cosine.

sin ax sin bx =12(cos (a − b)x − cos (a + b)x)

For example, the product of two sine functions can be converted into a single cosine function by taking the difference between two cosine functions.

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Therefore, integration of the left-hand side is equivalent to integrating two separate cosine functions, which can be done immediately.

sin ax cos bx =12(sin (a − b)x + sin (a + b)x)

Similarly, the product of a sine function and a cosine function can be written as the sum of two sine functions. Similarly, the product of two cosine functions can be written as the sum of two cosine functions.

cos ax cos bx =12(cos (a − b)x + cos (a + b)x)

Trigonometric integrals of the first type can be handled very easily by means of the compound angle formula.

For trigonometric functions involving parameters m and n, we recall that we saw reduction formulas for certain powers of sin x or cos x. This combination of two such functions makes our work slightly more complicated.

∫ sin2x dx = ∫1−cos 2x2dx

If we integrate sin x squared, we can use the reduction formula that we saw earlier in this chapter or directly use the double angle formula for cosine two x.

sin2x =1−cos 2x2

Thus we know that cosine two x is equal to one minus two sin x squared; using this formula, we can rewrite sin x squared as this expression and integrate it by integrating cosine two x.

∫ sin2x dx = ∫1−cos 2x2dx

=12∫(1 − cos 2x)dx

=12x −12sin 2x2 + C

=x2 −sin 2x4 + C

The final formula for the integration of cos x can be derived using the double-angle formula for cosine two x, yielding this result.

∫ cos2x dx = ∫1−cos 2x2dx cos2x =1−cos 2x2

=x2 +sin 2x4 + C

In general, we consider the integral of a combination of sin x and cosine x. m and n are nonnegative integers.

∫ sinmx cosnx dx (for integers m, n≥0)

We can divide the problem into three cases. In the first case, m is an odd number.

m is odd, m = 2k + 1

We can rewrite m as two k plus one. Then, using the identity sin x squared equals one minus cosine x squared, we have:

sinmx = sin2k+1x = (sin2x)ksin x = (1 − cos2x)ksin x

we let u = cos x, then sin x dx =− du, the integrand becomes (1 − u2)kdu

The equation sin m x equals this because m equals two k plus one. We take out one copy of sin x, leaving two k copies of sin x. Now the point in this separation is that we have a sin x square and we change it to one minus cosine x square. We have a sin x here, and then in case two, we have something very similar.

∫ sinmx cosnx dx (for integers m, n≥0)

m is even and n is odd. We write n = 2k + 1

In this case, we assume that n is odd. We can then write n as two times an integer plus one, or 2k + 1. We make use of this identity by transforming the given expression in accordance with it.

cosnx = cos2k+1x = (cos2x)kcos x = (1 − sin2x)kcos xlet u = sin x, then du = cos x dx, and the integrand becomes (1 − u2)kdu

If m and n are both even, this becomes cosine x to the two k plus one. We can take out one copy of cosine x and square it; two k becomes this one. We replace cosine x with this one.

∫ sinmx cosnx dx (for integers m, n≥0)