the domain of sec 5x is.
Answers
Since secant is 1/cosine, the domain must avoid division by 0:
5x ≠ (2n - 1)π/2 for n = …, -3, -2, -1, 0, 1, 2, 3, … because for these values, cosine is 0
Thus, x ≠(2n - 1)π/10 (having divided both sides by 5)
Let u = 5x. Thus, du = 5 dx and dx = du/5:
the_integral_in_terms_of_u
Write sec u = 1/cos u and multiply by (cos u / cos u):
multiplying_by_cos_over_cos
The denominator is cosine squared:
cos^2
From sin2u + cos2u = 1, write cos2u as 1 - sin2u:
cos^2_is_1-sin^2
The integration becomes:
integration_sec_u
Let v = sin u meaning dv = cos u du:
v=sin_u
Write 1 - v2 as (1 - v)(1 + v):
difference_of_squares
Do a partial fraction expansion:
PFE
Find A and B:
finding_A_and_B
1/(1 - v2) becomes:
PFE_with_A=B=1/2
The integration becomes:
integration_of_sec_u_with_PFE
Evaluating each integral on the right-hand side separately.
Let w = 1 - v, dw = -dv:
with_w=1-v
The integral of dw/w is ln |w|:
ln|w|
Substitute back 1 - v for w:
sub_1-v_for_w
Substitute back sin u for v:
sin_u_for_v
Substitute back 5x for u:
sub_5x_for_u
Evaluating the second integral, this time w = 1 + v and dw = dv:
with_w=1-v
As before:
ln|w|
Substitute back 1 + v for w:
sub_1+v_for_w
Substitute back sin u for v:
sin_u_for_v
Substitute back 5x for u:
sub_5x_for_u
The integral of sec 5x becomes:
with_the_two_integrals
The difference of logarithms is the logarithm of a quotient:
multiply_1/5_by_1/2
1/2 times 1/5 gives 1/10. An indefinite integral includes a constant, C.
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