Video: Evaluating the Definite Integral of a Function Involving Trigonometric Functions Using Integration by Substitution

Find ∫_(0) ^(πœ‹/4) βˆ’9 tan (𝑧) secΒ² (𝑧) d𝑧.

01:43

Video Transcript

Find the definite integral between zero and πœ‹ by four of negative nine tan 𝑧 times sec squared 𝑧 with respect to 𝑧.

Firstly, let’s not worry that this function is in terms of 𝑧. We’re integrating with respect to 𝑧, so we perform the process as normal. Then, we notice that sec squared 𝑧 is the derivative of tan of 𝑧. This tells us we can use integration by substitution to evaluate it. We’re going to let 𝑒 be equal to tan of 𝑧. And we know that the first derivative of tan of 𝑧 is sec squared 𝑧. And whilst d𝑒 by d𝑧 is not a fraction, we treat it a little like one, and we say that d𝑒 is equal to sec squared 𝑧 d𝑧.

And we now see that we can replace tan of 𝑧 with 𝑒 and we can replace sec squared 𝑧 d𝑧 with d𝑒. Before we go any further though, we’re going to need to work out what our new limits are. So, we use our substitution. We said 𝑒 is equal to tan of 𝑧 and our lower limit is when 𝑧 is equal to zero, which is when 𝑒 is equal to tan of zero, which is zero. Our upper limit is 𝑧 is equal to πœ‹ by four. So, 𝑒 is equal to tan of πœ‹ by four, which is one. And that’s great because we can now rewrite our definite integral as the definite integral between zero and one of negative nine 𝑒 with respect to 𝑒.

Now, we could, if we wanted, take negative nine out as a constant factor. Of course, we don’t need to. And if we don’t, we find that the integral of negative nine 𝑒 is negative nine 𝑒 squared divided by two. We’re going to evaluate this between zero and one. That’s negative nine times one squared over two minus negative nine times zero squared over two, which is just simply negative nine over two. Our definite integral is equal to negative nine over two.

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