Hi,
The question is worth 12 marks so the
working requires a considerable amount of mathematical
effort from the student.
It is important for you to appreciate why the
substitution was chosen for this question, which I will now
clarify.
We are given the expression sqrt [1 + e^(2x) ]. Whenever
we see a square root operation, we would like the expression inside
to be a perfect square to neutralise the square
root operation. We recall in trigonometry that 1 + (tan y)^2 =
(sec y)^2. Hence, we take the substitution to be tan y = e^x,
which is equivalent to x = ln (tan y) in Eagle's working.
This is another version:
Let tan y = e^x.
Differentiating with respect to x, (sec y)^2 .
(dy/dx) = e^x
=> dx/dy = (sec y)^2 / e^x = 1 / (cos y sin y),
since (sec y)^2 = 1 + (tan y)^2 and e^x = tan y.
Now int sqrt [ 1 + e^(2x) ] dx = int sqrt [ 1 + (tan
y)^2 ] (dx/dy) dy, where we use the chain rule (dx/dy) dy to change
the subject of integration from x to y.
= int (sec y) . {1 / (cos y sin y)} dy, noting that 1 /
(cos y sin y) = sec y cosec y.
= int (sec y)^2 . cosec y dy
= int {1 + (tan y)^2} . cosec y dy
= int {cosec y + (tan y)(sec y)} dy
= -ln |cosec y + cot y| + int { (sin y)(cos y)^(-2)
} dy
= -ln |cosec y + cot y| + sec y + c, which has to be
converted back to an expression involving x. We will need to find
the other trigonometric ratios with the aid of a right-angled
triangle.
Since tan y = e^x, we have
opposite side = e^x,
adjacent side = 1, and
hypothenuse side = sqrt [ 1 + e^(2x) ] in a
right-angled triangle, which you should draw to visualise.
From here, we can
find expressions for cosec y (ratio is
hypothenuse/opposite), cot y (ratio is adjacent/opposite) and sec y
(ratio is hypothenuse/adjacent), so that our final answer can be
represented in terms of x, as Eagle has provided.
Thanks.
Cheers,
Wen Shih