Video: Finding the Integration of a Function Involving a Trigonometric Function

Determine ∫(4π‘₯ βˆ’ 5 cos 6π‘₯) dπ‘₯.

02:35

Video Transcript

Determine the integral of four π‘₯ minus five times the cos of six π‘₯ with respect to π‘₯.

The question wants us to determine the integral of the difference between two functions. Our integrand is the difference between four π‘₯ and five times the cos of six π‘₯. And we know how to integrate each of these terms in our integrand separately. So, to determine this integral, we’ll just split our integral into two integrals which we can evaluate.

First, we recall that the integral of a difference between two functions is equal to the difference between their integrals. Using this, we can rewrite our integral as the integral of four π‘₯ with respect to π‘₯ minus the integral of five times the cos of six π‘₯ with respect to π‘₯. And we could now evaluate each of these integrals separately.

First, to integrate four π‘₯ with respect to π‘₯, we’ll use the power rule for integration. And we recall this tells us, for constants π‘Ž and 𝑛 where 𝑛 is not equal to negative one, to integrate π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯, we add one to our exponent and divide by this new exponent. Then, we add our constant of integration 𝑐. So, by thinking of four π‘₯ as four π‘₯ to the first power, we can evaluate our first integral as four π‘₯ squared divided by two plus the constant of integration we’ll call 𝑐 one.

We then need to evaluate our second integral. We can see that this is a trigonometric integral. And this is a standard trigonometric integral which we should commit to memory. For constants π‘Ž and 𝑛 where 𝑛 is not equal to zero, the integral of π‘Ž times the cos of 𝑛π‘₯ with respect to π‘₯ is equal to π‘Ž times the sin of 𝑛π‘₯ divided by 𝑛 plus the constant of integration 𝑐. So, by using our standard trigonometric integral rule, we get the integral of five times the cos of six π‘₯ with respect to π‘₯ is equal to five times the sin of six π‘₯ divided by six plus the constant of integration 𝑐 two.

And remember, since we were subtracting this integral, we need to subtract the result. Canceling the shared factor of two in the numerator and the denominator of four π‘₯ squared, distributing the negative over our parentheses, and rearranging, we get two π‘₯ squared minus five over six sin six π‘₯ plus 𝑐 one minus 𝑐 two.

And we can do one more thing to simplify this answer. Since 𝑐 one and negative 𝑐 two are both constants, we can combine both of these into a new constant, which we will call 𝑐. Therefore, we have shown the integral of four π‘₯ minus five times the cos of six π‘₯ with respect to π‘₯ is equal to two π‘₯ squared minus five over six times the sin of six π‘₯ plus 𝑐.

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