Manual How to understand Integral Calculus 5 Integration of products of sines and cosines

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Some of the following trigonometry identities may be needed. A.) $ \cos^2 x + PROBLEM 5: Integrate $ \displaystyle{ \int { 3 \cos^2 5x } \,. Click HERE to see a Click HERE to return to the original list of various types of calculus problems.
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So, we can use a similar technique in this integral. Of course, if both exponents are odd then we can use either method. Each integral is different and in some cases there will be more than one way to do the integral. With that being said most, if not all, of integrals involving products of sines and cosines in which both exponents are even can be done using one or more of the following formulas to rewrite the integrand.

The first two formulas are the standard half angle formula from a trig class written in a form that will be more convenient for us to use. The last is the standard double angle formula for sine, again with a small rewrite. As noted above there are often more than one way to do integrals in which both of the exponents are even. This integral is an example of that.

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There are at least two solution techniques for this problem. Solution 1 In this solution we will use the two half angle formulas above and just substitute them into the integral. In fact to eliminate the remaining problem term all that we need to do is reuse the first half angle formula given above.

Solution 2 In this solution we will use the half angle formula to help simplify the integral as follows.

In the previous example we saw two different solution methods that gave the same answer. Note that this will not always happen.

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In fact, more often than not we will get different answers. However, as we discussed in the Integration by Parts section, the two answers will differ by no more than a constant. Sometimes in the process of reducing integrals in which both exponents are even we will run across products of sine and cosine in which the arguments are different. These will require one of the following formulas to reduce the products to integrals that we can do.

Integral of sin(mt) and cos(mt)

The general integral will be,. Note that this method does require that we have at least one secant in the integral as well. If the exponent on the secant is even and the exponent on the tangent is odd then we can use either case. Again, it will be easier to convert the term with the smallest exponent.

Fourier series

That means that we need to strip out two secants and convert the rest to tangents. Here is the work for this integral. Both of the previous examples fit very nicely into the patterns discussed above and so were not all that difficult to work. However, there are a couple of exceptions to the patterns above and in these cases there is no single method that will work for every problem. Each integral will be different and may require different solution methods in order to evaluate the integral.


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To do this integral all we need to do is recall the definition of tangent in terms of sine and cosine and then this integral is nothing more than a Calculus I substitution. The simplification was done solely to eliminate the minus sign that was in front of the logarithm. This does not have to be done in general, but it is always easy to lose minus signs and in this case it was easy to eliminate it without introducing any real complexity to the answer and so we did.

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Note that all odd powers of tangent with the exception of the first power can be integrated using the same method we used in the previous example. However, if we manipulate the integrand as follows we can do it. The idea used in the above example is a nice idea to keep in mind.


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  • Multiplying the numerator and denominator of a term by the same term above can, on occasion, put the integral into a form that can be integrated. However, when it does work and you can figure out what term you need it can greatly simplify the integral.

    Trigonometric Integrals - Even Powers, Trig Identities, U-Substitution, Integration By Parts - Calcu

    Note that using integration by parts on this problem is not an obvious choice, but it does work very nicely here. The higher index integral can be used to calculate lower index ones; the process is continued repeatedly until we reach a point where the function to be integrated can be computed, usually when its index is 0 or 1.

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    Then we back-substitute the previous results until we have computed I n. The following integrals [3] contain:. The following integrals [4] contain:. From Wikipedia, the free encyclopedia. Part of a series of articles about Calculus Fundamental theorem Limits of functions Continuity. Mean value theorem Rolle's theorem. Differentiation notation Second derivative Third derivative Change of variables Implicit differentiation Related rates Taylor's theorem. Fractional Malliavin Stochastic Variations.

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