Video: Integrating a Function Involving Trigonometric Functions Using Integration by Substitution

Evaluate ∫_(πœ‹/6)^(πœ‹/2) 2 csc 𝑑 cot d𝑑.

03:03

Video Transcript

Evaluate the definite integral of two csc 𝑑 cot 𝑑 with the limits πœ‹ over two and πœ‹ over six.

So the first thing we’ve done is taken two outside of the integration because that’s just our constant. And then, what we’ve also done is swapped our cot 𝑑 and our csc 𝑑. So we have two multiplied by the definite integral of cot 𝑑 csc 𝑑 between the limits πœ‹ over two and πœ‹ over six.

So now, let’s take a look at how we deal with a definite integral. So if we want the definite integral of a function between limits 𝑏 and π‘Ž, then this is equal to the integral of that function with 𝑏 substituted in instead of π‘₯ minus the integral of that function with π‘Ž substituted in instead of π‘₯.

So now, what we can do is integrate cot 𝑑 csc 𝑑. And when we integrate this, we’re gonna get negative csc 𝑑 just because this is one of our standard integrals. So now, what we need to do is substitute in πœ‹ over two and πœ‹ over six for our 𝑑. So when we do that, we’ve got negative two multiplied by csc πœ‹ over two minus csc πœ‹ over six cause we’ve taken the negative out of the parentheses, which is gonna give us negative two multiplied by one minus two. And that’s cause the exact value of csc πœ‹ over two is one. And the exact value of csc πœ‹ over six is two.

And if you’re finding it hard to remember the exact values of csc πœ‹ over two and csc πœ‹ over six, we can think about it in terms of sine because csc π‘₯ is the same as one over sin π‘₯. So therefore, one over sin πœ‹ over two is gonna be equal to one over one, which is just one. And that’s because sin πœ‹ over two is an exact value we know, which is just one. Then remembering that if we have πœ‹ over two that’s in radians but it’s also sin 90 degrees if it was in degrees.

Okay, so let’s have a little look at csc πœ‹ over six. So for csc πœ‹ over six, it’s the same as one over sin πœ‹ over six. Well, this is equal to one over a half because the exact value for sin πœ‹ over six is a half, again remembering that πœ‹ over six is the same as 30 degrees if you want to think about it in degrees. And one divided by a half is just gonna be equal to two. Okay great. So we’ve just seen where they’ve come from.

So now, let’s get on and finish off the question. So we’ve got negative two multiplied by one minus two. Well, one minus two is just gonna be negative one. So negative two multiplied by negative one gives us two. So therefore, we can say that the definite integral of two csc 𝑑 cot 𝑑 between the limits πœ‹ over two and πœ‹ over six is equal to two.

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