Answers
EXPLANATION.
⇒ ∫1 - cos(2x)/1 + cos(2x) dx.
As we know that,
⇒ cos(2x) = 2cos²∅ - 1.
⇒ cos(2x) + 1 = 2cos²∅.
⇒ cos(2x) = 1 - 2sin²∅.
⇒ cos(2x) - 1 = -2sin²∅.
⇒ 1 - cos(2x) = 2sin²∅.
Substitute the value in equation, we get.
⇒ ∫1 - cos(2x)/1 + cos(2x)dx.
⇒ ∫2sin²x/2cos²x dx.
⇒ ∫tan²x.dx
Formula of tan²x,
⇒ tan²x = sec²x - 1.
Substitute the value in equation, we get.
⇒ ∫(sec²x - 1)dx.
⇒ ∫sec²x - ∫dx.
⇒ tan(x) - x + c.
MORE INFORMATION.
(1) = ∫sin(x)dx = -cos(x) + c.
(2) = ∫cos(x)dx = sin(x) + c.
(3) = ∫tan(x)dx = ㏒(sec(x)) + c Or -㏒(cos(x)) + c.
(4) = ∫cot(x)dx = ㏒(sin(x)) + c.
(5) = ∫sec(x)dx = ㏒(sec(x) + tan(x)) + c Or -㏒(sec(x) - tan(x)) + c Or ㏒tan(π/4 + x/2) + c.
(6) = ∫cosec(x)dx = -㏒(cosec(x) + cot(x)) + c Or ㏒(cosec(x) - cot(x) + c. Or ㏒tan(x/2) + c.
(7) = ∫sec(x).tan(x).dx = sec(x) + c.
(8) = ∫cosec(x).cot(x).dx = -cot(x) + c.
(9) = ∫sec²x.dx = tan(x) + c.
(10) = cosec²dx = -cot(x) + c.
I=integ.of (1+cos2x)/(1-cos2x) .dx
I =integ.of (1+2cos^2x-1)/(1–1+2sin^2x).dx
I=integ.of (2cos^2x)/(2sin^2x).dx
I=integ. of cot^2x.dx
I=integ.of (cosec^2x-1).dx
I= -cot x - x +C , Answer.
