Video: Integrating a Function by Recalling a Standard Trigonometric Derivative

Calculate ∫ 3/(4 + π‘₯Β²) dπ‘₯. [A] (4/3) tan⁻¹ (π‘₯) + 𝐢 [B] (3/2) tan⁻¹ (π‘₯/2) + 𝐢 [C] (3/2) tan⁻¹ (π‘₯) + 𝐢 [D] 3 tan⁻¹ (π‘₯/2) + 𝐢

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Video Transcript

Calculate the integral of three divided by four plus π‘₯ squared with respect to π‘₯. We’re given four options. a) four-thirds multiplied by the inverse tangent of π‘₯ plus our constant of integration 𝐢. b) three over two multiplied by the inverse tangent of π‘₯ divided by two plus our constant of integration 𝐢. c) three over two multiplied by the inverse tangent of π‘₯ plus our constant of integration 𝐢. And option d), three multiplied by the inverse tangent of π‘₯ divided by two plus the constant of integration 𝐢.

The question asks us to integrate three divided by four plus π‘₯ squared with respect to π‘₯. One thing we can do in any question asking us to integrate is to look at our integrand, that is the function inside of our integral, and to see if this reminds us of any standard derivatives that we might know. We can recall that the derivative with respect to π‘₯ of the inverse tangent of π‘₯ is equal to one divided by one plus π‘₯ squared. And the derivative with respect to π‘₯ of the inverse cotangent function of π‘₯ is equal to negative one divided by one plus π‘₯ squared. And we can see that these derivatives are very similar to the integrand given to us in the question.

What we can then do is integrate both sides of our equation to give us the integral of the derivative of the inverse tangent of π‘₯ with respect to π‘₯ is equal to the integral of one divided by one plus π‘₯ squared with respect to π‘₯. We then notice that the antiderivative of our derivative function of the inverse tangent of π‘₯ is just equal to the inverse tangent of π‘₯. And we add our constant of integration 𝐢. We could do the same process with the inverse cotangent of π‘₯ to get that the inverse cotangent of π‘₯ plus 𝐢 is equal to the integral of negative one divided by one plus π‘₯ squared with respect to π‘₯.

At this point, we could continue using either of these identities and arrive at a correct answer. However, the options listed in the question are all in terms of the inverse tangent function. So we will continue with using the inverse tangent function. So let’s start with our integral of three divided by four plus π‘₯ squared with respect to π‘₯. We’ve already shown we know how to integrate one divided by one plus π‘₯ squared with respect π‘₯. So let’s try and rearrange this to look like this integral.

Since the numerator of three is just a constant, we can take the constant factor of three outside of our integral. Next, we can take a factor of four outside of our denominator to get four multiplied by one plus π‘₯ squared divided by four. Then, we can notice that π‘₯ squared divided by four is actually equal to π‘₯ over two all squared. This gives us three multiplied by the integral of one divided by four multiplied by one plus π‘₯ over two all squared with respect to π‘₯. And similarly to how we did before, we could take the constant factor of four in our denominator outside of our integral. This gives us three-quarters multiplied by the integral of one divided by one plus π‘₯ over two all squared with respect to π‘₯.

We can now see that our integrand is very similar to one divided by one plus π‘₯ squared. The only difference is, instead of an π‘₯ squared, we have an π‘₯ over two squared. To rectify this, let’s use the substitution 𝑒 is equal to π‘₯ divided by two. This gives us that the derivative of 𝑒 with respect to π‘₯ is equal to a half. It’s worth noting here that d𝑒 dπ‘₯ is not actually a fraction. However, when using integration by substitution, it does behave a little bit like a fraction. This gives us that two d𝑒 is equal to dπ‘₯. Substituting this information into our integral gives us three quarters multiplied by the integral of one divided by one plus 𝑒 squared multiplied by two with respect to 𝑒. We could take the factor of two outside of our integral and then cancel it with the division by four.

We’re now ready to use our integral result. The integral of one divided by one plus 𝑒 squared with respect to 𝑒 is equal to the inverse tangent of 𝑒 plus a constant of integration 𝐢 one. Finally, we distribute three over two over our parentheses and use our substitution 𝑒 is equal to π‘₯ divided by two to get three over two multiplied by the inverse tangent of π‘₯ divided by two. And instead of adding three over two multiplied by 𝐢 one, we’ll add a new constant of integration which we will call 𝐢. Therefore, what we have shown is that the integral of three divided by four plus π‘₯ squared with respect to π‘₯ is equal to three over two multiplied by the inverse tangent of π‘₯ divided by two plus a constant of integration 𝐢, which was our option b.

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