Video: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions

Determine ∫ (βˆ’3 tanΒ² 8π‘₯ cscΒ² 8π‘₯) dπ‘₯.

01:29

Video Transcript

Determine the indefinite integral of negative three tan squared eight π‘₯ times csc squared eight π‘₯ evaluated with respect to π‘₯. These does at first look quite tricky. However, If we recall some of our trigonometry identities, it does get a little nicer. We know that tan π‘₯ is equal to sin π‘₯ over cos of π‘₯. And we also know that csc π‘₯ is equal to one over sin π‘₯. We can, therefore, rewrite our entire integrant as negative three times sin squared eight π‘₯ over cos squared eight π‘₯ times one over sin squared eight π‘₯. And then we noticed that the sin squared eight π‘₯ cancels. We can take the factor of negative three outside of the integral sin to make the next step easier. And we have a negative three times the integral of one over cos squared eight π‘₯ dπ‘₯.

But we know that one over cos of π‘₯ is equal to sec of π‘₯. So our integral becomes negative three times the integral of sec squared eight π‘₯ evaluated with respect to π‘₯. But of course, the integral of sec squared π‘Žπ‘₯, evaluated with respect to π‘₯, is one over π‘Ž tan of π‘Žπ‘₯ plus some constant of integration 𝑐. And so we see that the integral of sec squared eight π‘₯ is an eighth tan of eight π‘₯ plus 𝑐. We distribute our parentheses. And we see that we’re left with negative three-eights of tan of eight π‘₯ plus a new constant since we multiplied our original one by negative three. Let’s call that capital 𝐢.

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