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Question

Question: How do you find the definite integral for \[\cos \left( x \right)\] in the interval \[\left[ { - \pi...

How do you find the definite integral for cos(x)\cos \left( x \right) in the interval [π,π]\left[ { - \pi ,\pi } \right]?

Explanation

Solution

In the given question, we have been given a trigonometric function. We have to integrate it. To do that, we make the substitution of the integral using the standard results. Then we make substitutions and use the standard integration result, apply the formula of the definite integral using the limits to calculate the answer.

Formula Used:
We are going to use the formula of integration of cos(x)\cos \left( x \right), which is,
cos(x)dx=sin(x)\int {\cos \left( x \right)dx} = \sin \left( x \right)

Complete step by step answer:
We have to calculate the integral of cos(x)\cos \left( x \right).
I=cos(x)dxI = \int {\cos \left( x \right)dx}
We know,
d(sinx+c)dx=cosx\dfrac{{d\left( {\sin x + c} \right)}}{{dx}} = \cos x
Hence, we have,
cos(x)dx=sinx+c\int {\cos \left( x \right)dx} = \sin x + c
Now we put in the value of the limits,

ππcos(x)dx=[sinx]+ππ=sin(π)sin(π)=sinπ+sinπ=0\int\limits_{ - \pi }^\pi {\cos \left( x \right)dx} = {\left[ {\sin x} \right]^{ + \pi }}_{ - \pi } = \sin \left( \pi \right) - \sin \left( { - \pi } \right) = \sin \pi + \sin \pi = 0

Additional Information:
Integration is the opposite of differentiation. In differentiation, we “break” things for examining how they behave separately. While, in integration, we combine the expressions so as to see their collective behavior. If we have a definite integral, then we calculate its value by putting in the upper limit into the result, then putting in the lower limit into the result, and then subtracting the two. A definite integral is the one which looks like,
basomeexpression\int\limits_b^a {some{\rm{ }}\exp ression} .

Note:
In the given question, we had to find the integral of cosine. We did that by using the standard result to evaluate the value of the expression. Then we put in the values of the limits and evaluated the answer.