Math, asked by rimagoswami49, 3 months ago

inverse dx /
1 + sinx​

Answers

Answered by amansharma264
10

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.


pulakmath007: Excellent Aman !!
Answered by xXMarziyaXx
0

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.

#Be brainly

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