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Question Video: Finding the Equation of a Curve given the Expression of the Slope of Its Tangent Using Integration and Trigonometric Identities Mathematics

Find the equation of the curve given the gradient of the tangent is 5 sinΒ² (π‘₯/2) and the curve passes through the origin.

04:18

Video Transcript

Find the equation of the curve given the gradient of the tangent is five sin squared π‘₯ over two and the curve passes through the origin.

Remember, if we know the equation of a curve, we can find the value of its gradient or slope at a point by differentiating. We also know that integration is the reverse process to differentiation. And so, we’re going to be able to find a general equation of our curve by integrating our expression for the gradient of the tangent. That’s integrating five sin squared π‘₯ over two with respect to π‘₯. We’re told that the curve passes through the origin. Remember, that’s the point whose coordinates are zero, zero. So, we’ll use this information in a little while to find the particular solution or the particular equation of the curve.

Let’s start then by integrating five sin squared π‘₯ over two with respect to π‘₯. Now, whilst it’s not entirely necessary, it can be nice to remove any constant factors and focus on integrating the function itself. So, we can write this as five times the integral of sin squared π‘₯ over two. But how do we integrate sin squared π‘₯ over two? Well, we’re going to begin by recalling one of our double-angle formulae.

We know that cos of two πœƒ is cos squared πœƒ minus sin squared πœƒ. But we also know that cos squared πœƒ plus sin squared πœƒ is equal to one. Sometimes, we use identity symbols instead of equal symbols here. And this is because these equations or identities are true for all values of πœƒ, but we’re just going to use equal signs and bear this in mind.

From our second identity, we’re going to subtract sin squared πœƒ from both sides. That tells us that cos squared πœƒ is equal to one minus sin squared πœƒ. Then, we replace cos squared πœƒ with one minus sin squared πœƒ in our first identity. And we find that cos two πœƒ is one minus sin squared πœƒ minus sin squared πœƒ or cos two πœƒ is one minus two sin squared πœƒ. Because we’re integrating an expression in sin squared, we’re going to rearrange to make sin squared πœƒ the subject.

We’ll add two sin squared πœƒ to both sides and subtract cos two πœƒ. So, two sin squared πœƒ is one minus cos two πœƒ. And if we divide through by two, we find sin squared πœƒ is one minus cos two πœƒ all over two. Now, this looks a little bit different to our expression for sin squared π‘₯ over two. But remember, we said these identities are true for all values of πœƒ. So, let’s see what happens if we let πœƒ be equal to π‘₯ over two.

When we do, we see that sin squared of π‘₯ over two is equal to one minus cos of two times π‘₯ over two all over two. But two times π‘₯ over two is simply π‘₯. So, we found that sin squared of π‘₯ over two is equal to one minus cos π‘₯ over two. And we now see that to integrate sin squared π‘₯ over two, we actually integrate one minus cos π‘₯ over two with respect to π‘₯. Once again, we can take out constant factors. Let’s take out one-half. And we see that 𝑦 is going to be equal to five over two times the integral of one minus cos of π‘₯ with respect to π‘₯. Let’s integrate term by term.

When we integrate one with respect to π‘₯, we simply get π‘₯. We know that the integral of cos of π‘₯ is sin π‘₯. So, when we integrate negative cos of π‘₯, we get negative sin π‘₯. Of course, this is an indefinite integral, so we need to take into account that there’s going to be a constant of integration. Let’s call that 𝐴. Now, this step isn’t entirely necessary. We could do this later, but we might choose at this stage to distribute our parentheses. When we do, we get 𝑦 is equal to five over two π‘₯ minus five over two sin π‘₯ plus 𝐡. This is essentially a new constant, since we’ve multiplied our earlier constant by five over two.

Now, of course, we know that our curve passes through the origin. That’s the point with coordinates zero, zero. In other words, there is a point on our line that when π‘₯ is equal to zero, 𝑦 is equal to zero. So, let’s replace π‘₯ and 𝑦 in our equation. It becomes zero is equal to five over two times zero minus five over two times sin zero plus 𝐡. Five over two times zero is zero, and five over two times sin zero is also zero. So, 𝐡, our constant, is actually equal to zero.

And so, we’ve found the equation of the curve. It’s 𝑦 equals five over two π‘₯ minus five over two sin π‘₯.

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