x-sinx-1-cosx-dx-please-help-me-

Question Number 154195 by rexford last updated on 15/Sep/21

\int \frac{x + s i n x}{1 + c o s x} d x

p l e a s e, h e l p m e

Answered by qaz last updated on 15/Sep/21

∵ {(\frac{\sin x}{1 + \cos x})}^{'} = \frac{1}{1 + \cos x}

∴ \int \frac{x + \sin x}{1 + \cos x} dx

= \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx

= \int xd (\frac{\sin x}{1 + \cos x}) + \int \frac{\sin x}{1 + \cos x} dx

= \frac{xsin x}{1 + \cos x} + C

Commented by qaz last updated on 15/Sep/21

\int \frac{x - \sin x}{1 + \cos x} dx \dots \dots \int \frac{x - \sin x}{1 - \cos x} dx \dots \dots \int \frac{x + \cos x}{1 + \sin x} dx \dots . . \int \frac{x - \cos x}{1 + \sin x} dx

\int \frac{x + \cos x}{1 - \sin x} dx \dots \dots \int \frac{x - \cos x}{1 - \sin x} dx \dots \dots . \int \frac{x + \sin x}{1 - \cos x} dx

The same \dots .

Answered by puissant last updated on 15/Sep/21

Ω = \int \frac{x + s i n x}{1 + c o s x} d x

= \int \frac{x}{1 + c o s x} d x + \int \frac{s i n x}{1 + c o s x} d x

= \int {x s e c}^{2} (\frac{x}{2}) d x - l n ∣ c o s x ∣ + C

= x t a n (\frac{x}{2}) - \int t a n (\frac{x}{2}) d x - l n ∣ c o s x ∣ + C

= x t a n (\frac{x}{2}) + 2 l n ∣ c o s (\frac{x}{2}) ∣ - l n ∣ c o s x ∣ + C

∴∵ Ω = x t a n (\frac{x}{2}) + 2 l n ∣ c o s (\frac{x}{2}) ∣ - l n ∣ c o s x ∣ + C . .

Answered by maged last updated on 15/Sep/21

I = \int \frac{x}{1 + \cos x} . \frac{1 - \cos x}{1 - \cos x} d x + \int \frac{\sin x}{1 + \cos x} . \frac{1 - \cos x}{1 - \cos x} d x

= \int \frac{x (1 - \cos x)}{\sin^{2} x} d x + \int \frac{1}{\sin x} d x

I = \int x {cosec}^{2} x d x - \int x \cot x cosec x d x + \int cosec x d x

u = x \to d u = d x

d v = {cosec}^{2} x d x \to v = - \cot x

- -

u = x \to d u = d x

d v = - \cot x cosec x d x \to v = cosec x

- -

I = - x \cot x + x cosec x + \int \cot x d x - \int cosec x d x + \int cosec x d x

= x cosec x - x \cot x - \ln (\sin x) + C

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