Question Video: Integrating Trigonometric Functions | Nagwa Question Video: Integrating Trigonometric Functions | Nagwa

# Question Video: Integrating Trigonometric Functions Mathematics • Higher Education

Suppose that dπ¦/dπ₯ = β9 sin 2π₯ β 3 cos 5π₯ and π¦ = 7 when π₯ = π/6. Find π¦ in terms of π₯.

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### Video Transcript

Suppose that dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Find π¦ in terms of π₯.

The question gives us a first-order differential equation. It wants us to find π¦ in terms of π₯ given that when π¦ is equal to seven π₯ is equal to π by six. We recall if weβre told the derivative of π¦ with respect to π₯ is equal to some function of π₯, we can solve for π¦ by doing the opposite of differentiating. We have that π¦ is equal to the integral of dπ¦ by dπ₯ with respect to π₯, which we know is equal to the integral of negative nine times the sin of two π₯ minus three times the cos of five π₯ with respect to π₯.

To evaluate this integral, we can integrate each term separately. And we recall for constants π and π where π is not equal to zero, the integral of π cos of ππ₯ with respect to π₯ is equal to π times the sin of ππ₯ divided by π plus a constant of integration πΆ. And the integral of π sin of ππ₯ with respect to π₯ is equal to negative π the cos of ππ₯ divided by π plus the constant of integration πΆ.

Using this, we get the integral of negative nine times the sin of two π₯ is negative one multiplied by negative nine cos of two π₯ divided by two. And negative one multiplied by negative one is just equal to one. So, integrating the first term, we get nine cos of two π₯ divided by two. We can then do the same to evaluate the integral of negative three times the cos of five π₯. We get negative three times the sin of five π₯ divided by five. And then, we add our constant of integration πΆ. And itβs worth noting we only need to add one constant of integration.

So, weβve now shown that π¦ is equal to this function of π₯, where πΆ can be any constant. However, weβre told that when π¦ is equal to seven, π₯ is equal to π by six. We can use this information to find the specific value of πΆ in this case. We substitute π¦ is equal to seven and π₯ is equal to π by six into this equation. This gives us that seven is equal to nine times the cos of two times π by six over two minus three times the sin of five times π by six over five plus πΆ.

We can simplify this. We have two times π by six is equal to π by three. Weβre now in a position to evaluate this expression. The cos of π by three is a standard trigonometric result which we should know. Itβs equal to one-half. And the sin of five π by six is also a standard trigonometric result we should know. Itβs also equal to one-half. This gives us that seven is equal to nine times one-half over two minus three times one-half over five plus πΆ. We can simplify this further. Nine times one-half over two is the same as saying nine over four. And three times one-half over five is the same as saying three over 10.

So, we have that seven is equal to nine over four minus three over 10 plus πΆ. We can rearrange this expression and then evaluate to get that πΆ is equal to 101 divided by 20. We then substitute this value of πΆ into our general solution. And this gives us that π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20. Therefore, weβve shown if dπ¦ by dπ₯ is equal to negative nine times the sin of two π₯ minus three times the cos of five π₯ and π¦ is equal to seven when π₯ is equal to π by six. Then π¦ is equal to negative three over five times the sin of five π₯ plus nine over two times the cos of two π₯ plus 101 divided by 20.

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