Video Transcript
Find the definite integral between
zero and 𝜋 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 𝑢 be equal to
tan of 𝑧. And we know that the first
derivative of tan of 𝑧 is sec squared 𝑧. And whilst d𝑢 by d𝑧 is not a
fraction, we treat it a little like one, and we say that d𝑢 is equal to sec squared
𝑧 d𝑧.
And we now see that we can replace
tan of 𝑧 with 𝑢 and we can replace sec squared 𝑧 d𝑧 with d𝑢. 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 𝑢 is equal to tan of 𝑧
and our lower limit is when 𝑧 is equal to zero, which is when 𝑢 is equal to tan of
zero, which is zero. Our upper limit is 𝑧 is equal to
𝜋 by four. So, 𝑢 is equal to tan of 𝜋 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 𝑢 with respect to 𝑢.
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 𝑢 is negative nine 𝑢 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.
