Question

integration cosx/1+cosx​

Answer 1

Solution :-

$\rm \mapsto \displaystyle \int\bigg(\dfrac{cosx}{1 + cosx} \bigg)dx.$

As we know that,

• Multiply and divide both denominator and numerator by 1, we get.

$\rm \mapsto \displaystyle \int \bigg(\dfrac{cosx + 1 - 1}{1 + cosx} \bigg)dx$

$\rm \mapsto \displaystyle \int \bigg(\dfrac{cosx + 1}{1 + cosx} \bigg)dx \ - \ \int \bigg(\dfrac{1}{1 + cosx} \bigg)dx.$

$\rm \mapsto \displaystyle \int dx \ - \ \int \bigg(\dfrac{1}{1 + cosx} \bigg)dx$

As we know that,

Formula of :

⇒ cos2∅  = 2cos²∅ - 1.

⇒ cos2∅ + 1 = 2cos²∅.

As we know that,

Put the value of ∅ = x/2 in equation, we get.

⇒ cos2(x/2) + 1 = 2cos²(x/2).

⇒ cos(x) + 1 = 2cos²x/2.

Put the value in the equation, we get.

$\rm \mapsto \displaystyle \int dx \ - \ \int \bigg(\frac{1}{2cos^{2}\dfrac{x}{2} } \bigg)dx$

$\rm \mapsto \displaystyle \int dx \ - \ \dfrac{1}{2} \int sec^{2} \frac{x}{2} dx$

$\rm \mapsto x \ - \ \dfrac{1}{2} \bigg[\dfrac{tan\dfrac{x}{2} }{\dfrac{1}{2} } \bigg] + c.$

$\rm \mapsto x - tan\dfrac{x}{2} + c.$

$\rm \mapsto \displaystyle \int\bigg(\dfrac{cosx}{1 + cosx} \bigg)dx. = x - tan\dfrac{x}{2} + c.$

                                                                                                                         

More to Know,

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.

Answer 2

Answer:

$→\int \frac{ \cos(x) }{1 + \cos(x) } dx \\ = \int \frac{ \cos(x) + 1 - 1 }{1 + \cos(x) } dx \\ = \int \frac{1 + \cos(x) }{1 + cos(x)} dx - \int \frac{1}{1 + \cos(x) } dx \\ = \int dx - \int \frac{1}{2 { \cos}^{2}( \frac{x}{2} ) } dx \\ = \int dx- \frac{1}{2} \int \sec ^{2} ( \frac{x}{2} ) dx \\ = x - \frac{1}{2} ( \frac{ \tan( \frac{x}{2} ) }{ \frac{1}{2} } ) + C \\ = x - \tan( \frac{x}{2} ) + C$

• x-tan(x/2)+C is the right answer.