inverse dx /
1 + sinx
Answers
EXPLANATION.
⇒ ∫dx/1 + sin(x).
As we know that,
Rationalize the equation, we get.
⇒ ∫1/1 + sin(x) X 1 - sin(x)/1 - sin(x)dx.
⇒ ∫(1 - sin(x))dx/(1 - sin²x).
As we know that,
Formula of :
⇒ sin²x + cos²x = 1.
⇒ cos²x = 1 - sin²x.
Put the values in the equation, we get.
⇒ ∫(1 - sin(x))dx/cos²x.
⇒ ∫1.dx/cos²x - ∫sin(x).dx/cos²x.
⇒ ∫sec²x.dx - ∫tan(x).sec(x)dx.
⇒ ∫(sec²x - tan(x).sec(x)).dx.
⇒ tan(x) - sec(x) + c.
MORE INFORMATION.
Some standard integrals.
(1) = ∫0.dx = c.
(2) = ∫1.dx = x + c.
(3) = ∫k dx = kx + c, ( k∈ R).
(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, ( n ≠ -1).
(5) = ∫dx/x = ㏒(x) + c.
(6) = ∫eˣdx = eˣ + c.
(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.
EXPLANATION.
⇒ ∫dx/1 + sin(x).
As we know that,
Rationalize the equation, we get.
⇒ ∫1/1 + sin(x) X 1 - sin(x)/1 - sin(x)dx.
⇒ ∫(1 - sin(x))dx/(1 - sin²x).
As we know that,
As we know that,Formula of :
⇒ sin²x + cos²x = 1.
⇒ cos²x = 1 - sin²x.
Put the values in the equation, we get.
⇒ ∫(1 - sin(x))dx/cos²x.
⇒ ∫1.dx/cos²x - ∫sin(x).dx/cos²x.
⇒ ∫sec²x.dx - ∫tan(x).sec(x)dx.
⇒ ∫(sec²x - tan(x).sec(x)).dx.
⇒ tan(x) - sec(x) + c.
MORE INFORMATION.
Some standard integrals.
(1) = ∫0.dx = c.
(2) = ∫1.dx = x + c.
(3) = ∫k dx = kx + c, ( k∈ R).
(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, ( n ≠ -1).
(5) = ∫dx/x = ㏒(x) + c.
(6) = ∫eˣdx = eˣ + c.
(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.