### Video Transcript

The slope at the point π₯, π¦ on a graph of a function is dπ¦ by dπ₯ equals negative four π sin of ππ₯ plus five π cos of ππ₯. Find the equation of the curve if it contains the point one, two.

In this question, weβve been given information about the derivative of our original function. Now, weβre looking to find the equation of the curve, given some information about a point that it passes through. In other words, we need to establish an equation for π¦ in terms of π₯. And so we begin by recalling that the fundamental theorem of calculus tells us that integration and differentiation are essentially reverse processes. And so weβll find an equation for π¦ by integrating our expression for dπ¦ by dπ₯ with respect to π₯.

In this case, π¦ is then the indefinite integral of negative four π sin of ππ₯ plus five π cos of ππ₯ with respect to π₯. And actually, we can quote the general result for the integral of the sine and cosine functions. The integral of sin of ππ₯ for some real constants π is negative one over π cos of ππ₯ plus a constant of integration π. Similarly, integrating cos of ππ₯ with respect to π₯ is one over π times sin of ππ₯ plus π. Remember, we can integrate term by term. So we begin by integrating negative four π of sin ππ₯ with respect to π₯. Itβs negative four π times negative one over π cos of ππ₯, which is four cos of ππ₯.

Then when we integrate five π cos of ππ₯, we get five π times one over π sin of ππ₯, which is five sin of ππ₯. And since weβre performing an indefinite integral, itβs really important that we have that constant of integration π. So weβve found that π¦ is equal to four cos of ππ₯ plus five sin of ππ₯ plus π. Now, we really need to find the value of π. So we go back to the question and the information that the curve passes through the point one, two. In other words, when π₯ is equal to one, π¦ is equal to two. And we substitute these values into our equation. We get two equals four cos of π times one plus five sin of π times one plus π.

Now, we know that five sin of π is zero. Similarly, cos of π is negative one. So four cos of π is negative four. And our equation becomes two equals negative four plus π, which we can solve for π by adding four to both sides. So π is equal to six. And we can, therefore, replace π with six in our equation for π¦. Thatβs π¦ equals four cos of ππ₯ plus five sin of ππ₯ plus six. Or if we choose by convention to write summing first, we get π¦ equals five sin of ππ₯ plus four cos of ππ₯ plus six.