Video: Integrating Sine Squared Using Half Angle Trigonometric Formulae

Determine ∫ (7 sinΒ² (5π‘₯) + 7) dπ‘₯.

03:44

Video Transcript

Determine the indefinite integral of seven sin squared of five π‘₯ plus seven with respect to π‘₯.

The first thing we notice here is that we have a factor of seven inside the interval, which can be factored out of the integral. This leaves us with seven times the indefinite integral of sin squared of five π‘₯ plus one with respect to π‘₯.

At this stage, we’re unable to integrate since we have a sin squared of five π‘₯ in our integral. In order to get rid of this square in our integral, we can use the cos double angle formula. This tells us that cos of two πœƒ is equal to one minus two sin squared πœƒ.

And now we can rearrange this formula for sin squared πœƒ, which gives us sin squared πœƒ is equal to one minus cos of two πœƒ over two. And now we can compare this to what we have in our integral. And we see that we have sin squared of five π‘₯ instead of sin squared of πœƒ. So let’s substitute in πœƒ equals five π‘₯. And we get that sin squared of five π‘₯ is equal to one minus cos of 10π‘₯ all over two.

And now we can substitute this into our integral. We have seven times the indefinite integral of one minus cos of 10π‘₯ all over two plus one with respect to π‘₯. And now we know that we can write one as two over two, and so we can incorporate it into our fraction. That gives us seven times the indefinite integral of three minus cos of 10π‘₯ all over two with respect to π‘₯.

And now we have a common factor of one-half inside our integral. So we can factor this out of our integral, leaving us with seven over two times the indefinite integral of three minus cos of 10π‘₯ with respect to π‘₯.

And now we’re ready to integrate this. So the constant in front of our integral remains the same. And for the first term in our integral, which is three, we can write that three is equal to three times π‘₯ to the naught, since π‘₯ to the naught is just equal to one. And now when we integrate three, we increase the power by one, so we get three timesed by π‘₯ to the one. And we divide by the new power, which is just one.

And our next term in the integral is minus cos of 10π‘₯. In order to integrate this, we remember that the integral of cos of π‘₯ with respect to π‘₯ is equal to sin of π‘₯. And so the integral of minus cos of 10π‘₯ will be minus sin of 10π‘₯. However, we mustn’t forget to use the inverse chain rule here, which means that we must divide by the derivative of the inside of our function, so that’s the derivative of 10π‘₯ with respect to π‘₯. So the derivative of 10π‘₯ is just 10, so we multiply our sin of 10π‘₯ term by one over 10. And now we have completed the integration.

However, since this was an indefinite integral, we mustn’t forget to add on our constant term, which is 𝑐, which is also called the constant of integration. And hence, we obtain a final solution that the indefinite integral of seven sin squared of five π‘₯ plus seven with respect to π‘₯ is equal to 21 over two π‘₯ minus seven over 20 sin of 10 π‘₯ plus 𝑐.

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