∫ [sin(x) + x] / [cos(x) + 1] dx = ?

Solution:

We will be solving this by using trigonometric/hyperbolic identities.

Let's solve this step by step.

Simplify the equation:

∫ [sin(x) + x] / [cos(x) + 1] dx = ∫ [sin(x) / {cos(x) + 1} + x / {cos(x) + 1}] dx

= ∫ sin(x) / {cos(x) + 1} dx + ∫ x / {cos(x) + 1} dx

Let us solve first ∫ sin(x) / {cos(x) + 1} dx 

Substitute z = {cos(x) + 1}

⇒ dz/dx = -sin(x)

sin(x) dx = -dz

Substitue sinx dx value in ∫ sin(x) / {cos(x) + 1} dx

∫ sin(x) / {cos(x) + 1} dx = - ∫ 1/z dz

= - ln(z)

Substitue z = {cos(x) + 1}; we get,

∫ sin(x) / {cos(x) + 1} dx  = - ln{cos(x) + 1}

Now let us solve the second part ∫ x / {cos(x) + 1} dx 

Multipy and divide by csc(x):

∫ x / {cos(x) + 1} dx = ∫ x csc(x) / {csc(x) + cot(x)} dx

Let's integrate this by parts: ∫ f g′ = f g − ∫ f′ g

Here f = x and g' = csc(x) / {csc(x) + cot(x)}

⇒ f′ = 1 and g = 1 / {csc(x) + cot(x)}

⇒ ∫ x / {cos(x) + 1} dx =  x / {csc(x) + cot(x)} - ∫ 1 / {csc(x) + cot(x)} dx

Now let us solve ∫ 1 / {csc(x) + cot(x)} dx 

Substitue csc(x) = 1 / sin(x) and cot(x) = cos(x) / sin(x) above:

∫ 1 / {csc(x) + cot(x)} dx = ∫ -[ -1 / {cos(x) + 1}] sin(x) dx

Substitue u = cos(x),

du/dx = -sin(x)

sin(x) dx = -du

∫ 1 / {csc(x) + cot(x)} dx = ∫ -[ -1 / {cos(x) + 1}] sin(x) dx  = ∫ -1 /(u + 1) du

Substitue v = u + 1

dv/du = 1

du = dv

 ∫ -1 /(u + 1) du  =  ∫ -1 /v dv = - ln(v)  = - ln(u + 1) = - ln[cos(x) + 1]

⇒ ∫ 1 / {csc(x) + cot(x)} dx = - ln[cos(x) + 1]

∫ x / {cos(x) + 1} dx =  x / {csc(x) + cot(x)} - ∫ 1 / {csc(x) + cot(x)} dx 

Substitue the value of ∫ 1 / {csc(x) + cot(x)} dx in the above equation:

∫ x / {cos(x) + 1} dx =  x / {csc(x) + cot(x)} + ln[cos(x) + 1]

Therefore, combining both parts

∫ [sin(x) + x] / [cos(x) + 1] dx = - ln{cos(x) + 1} + x / {csc(x) + cot(x)} + ln[cos(x) + 1] + C

= x / [csc(x) + cot(x)] + C

Hence,  ∫ [sin(x) + x] / [cos(x) + 1] dx = x / [csc(x) + cot(x)] + C

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∫ [sin(x) + x] / [cos(x) + 1] dx = ?

Summary:

Answer: ∫ [sin(x) + x] / [cos(x) + 1] dx = x / [csc(x) + cot(x)] + C