In class we saw that tan x + sec x dx tan x + sec x SEC x dx sec x · sec x tan x + sec2 x dx sec x + tan x du и In sec x + tan x + C where we used the u-substitution u = sec x + tan x. In this problem, you will derive the antiderivative of f(x) = sec x in another way. (a) Observe that COS x COS x cos? x COS x 1– sin? x COS x COs x 2 |1+ sin x 1– sin x 1+ sin x Show that SEC x dx In + C. [Hint: In(a) = ln(Va)] 1– sin x (b) Continuing from (a), show that f sec x dx = ln | sec x + tan x| + C. [Hint: You can use the multiply by 1 trick.] (c) What about this method is inspired from partial fractions?
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Hello! How do you solve part (a), (b), and (c) of the question that is attached?
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![In class we saw that
tan x + sec x
dx
tan x + sec x
SEC x dx
sec x ·
sec x tan x + sec2 x
dx
sec x + tan x
du
и
In sec x + tan x + C
where we used the u-substitution u = sec x + tan x. In this problem, you will derive the
antiderivative of f(x)
= sec x in another way.
(a) Observe that
COS x
COS x
cos? x
COS x
1– sin? x
COS x
COs x
2 |1+ sin x
1– sin x
1+ sin x
Show that
SEC x dx
In
+ C. [Hint: In(a) = ln(Va)]
1– sin x
(b) Continuing from (a), show that f sec x dx = ln | sec x + tan x| + C. [Hint: You can use
the multiply by 1 trick.]
(c) What about this method is inspired from partial fractions?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11b92d9b-5da5-4fb6-9df0-b83dcd17a243%2Fb8bb9328-d150-4918-bb0c-3cad9101cbe0%2Fu2hy734.png&w=3840&q=75)
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