1-1-cot-x-dx-

Question Number 5838 by gourav~ last updated on 31/May/16

\int \frac{1}{1 + \cot x} d x

Commented by Yozzii last updated on 31/May/16

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

L e t t = t a n 0.5 x \Rightarrow d t = 0.5 {s e c}^{2} 0.5 d x

d x = \frac{2}{1 + t^{2}} d t (1 + {t a n}^{2} 0.5 x = {s e c}^{2} . 5 x)

c o s x = \frac{1 - t^{2}}{1 + t^{2}}, s i n x = \frac{2 t}{1 + t^{2}}

∴ \int \frac{1}{1 + c o t x} d x = \int \frac{2 t \times 2}{1 - t^{2} + 2 t} d t

= - 2 \int \frac{- 2 + 2 - 2 t}{1 + 2 t - t^{2}} d t

= - 2 {\int \frac{\frac{d}{d t} (1 + 2 t - t^{2})}{1 + 2 t - t^{2}} d t - 2 \int \frac{d t}{2 - {(t - 1)}^{2}}}

\int \frac{d x}{1 + c o t x} = - 2 l n ∣ 1 + 2 t - t^{2} ∣ + 4 \int \frac{d t}{(\sqrt{2} - t + 1) (\sqrt{2} + t - 1)}

\frac{1}{\sqrt{2} + 1 - t} + \frac{1}{\sqrt{2} - 1 + t} = \frac{\sqrt{2} - 1 + t + \sqrt{2} + 1 - t}{2 - {(t - 1)}^{2}}

= \frac{2 \sqrt{2}}{2 - {(t - 1)}^{2}}

\Rightarrow \frac{1}{2 \sqrt{2}} (\frac{1}{t + \sqrt{2} - 1} + \frac{1}{\sqrt{2} + 1 - t}) = \frac{1}{(\sqrt{2} - 1 + t) (\sqrt{2} + 1 - t)}

∴ \int \frac{d x}{1 + c o t x} = \sqrt{2} \int (\frac{1}{t + \sqrt{2} - 1} - \frac{1}{- \sqrt{2} - 1 + t}) d t + l n (\frac{1}{{(1 + 2 t - t^{2})}^{2}})

\int \frac{d x}{1 + c o t x} = \sqrt{2} l n ∣ \frac{t - 1 + \sqrt{2}}{t - 1 - \sqrt{2}} ∣ + l n (\frac{1}{{(1 + 2 t - t^{2})}^{2}}) + C

\int \frac{d x}{1 + c o t x} = l n {\frac{∣ t a n 0.5 x - 1 + \sqrt{2} ∣^{\sqrt{2}}}{∣ t a n 0.5 x - 1 - \sqrt{2} ∣^{\sqrt{2}} {(1 + 2 t a n 0.5 x - {t a n}^{2} 0.5 x)}^{2}}} + C

\int \frac{d x}{1 + c o t x} = \sqrt{2} l n ∣ t - 1 + \sqrt{2} ∣ - \sqrt{2} l n ∣ t - 1 - \sqrt{2} ∣ - 2 l n ∣ t - 1 + \sqrt{2} ∣ - 2 l n ∣ t - 1 - \sqrt{2} ∣ + C

\int \frac{d x}{1 + c o t x} = (\sqrt{2} - 2) l n ∣ t - 1 + \sqrt{2} ∣ - (\sqrt{2} + 2) l n ∣ t - 1 - \sqrt{2} ∣ + C

\int \frac{d x}{1 + c o t x} = l n ∣ \frac{1}{∣ t a n 0.5 x - 1 + \sqrt{2} ∣^{2 - \sqrt{2}} ∣ t a n 0.5 x - 1 - \sqrt{2} ∣^{2 + \sqrt{2}}} ∣ + C

Commented by Yozzii last updated on 31/May/16

L e t R = \int \frac{s i n x}{c o s x + s i n x} d x a n d J = \int \frac{c o s x}{c o s x + s i n x} d x .

∴ R + J = \int 1 d x = x + C

J - R = \int \frac{c o s x - s i n x}{c o s x + s i n x} d x = \int \frac{\frac{d}{d x} (c o s x + s i n x)}{c o s x + s i n x} d x

J - R = l n ∣ c o s x + s i n x ∣ + N

∴ R + J - J + R = x - l n ∣ c o s x + s i n x ∣ + C - N

R = 0.5 (x - l n ∣ c o s x + s i n x ∣) + P

\int \frac{d x}{1 + c o t x} = \frac{1}{2} (x - l n ∣ c o s x + s i n x ∣) + P

P = c o n s t a n t o f i n t e g r a t i o n .

Commented by gourav~ last updated on 01/Jun/16

t h a n k y o u . = .

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