find-dx-1-cos-x-cos-2x-

Question Number 47059 by maxmathsup by imad last updated on 04/Nov/18

f i n d \int \frac{d x}{1 + c o s x + c o s (2 x)}

Commented by maxmathsup by imad last updated on 06/Nov/18

l e t A = \int \frac{d x}{1 + c o s x + c o s (2 x)} = \int \frac{d x}{1 + c o s (x) + 2 {c o s}^{2} x - 1}

= \int \frac{d x}{c o s x (1 + 2 c o s x)} = \int (\frac{1}{c o s x} - \frac{2}{1 + 2 c o s x}) d x = \int \frac{d x}{c o s x} - \int \frac{2}{1 + 2 c o s x} d x

b u t \int \frac{d x}{c o s x} =_{t a n (\frac{x}{2}) = t} \int \frac{1}{\frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{1 - t^{2}} = \int (\frac{1}{1 - t} + \frac{1}{1 + t}) d t

= l n ∣ \frac{1 + t}{1 - t} ∣ + c_{1} = l n ∣ \frac{1 + t a n (\frac{x}{2})}{1 - t a n (\frac{x}{2})} ∣= l n ∣ t a n (\frac{x}{2} + \frac{π}{4}) ∣ + c_{1} a l s o w e h a v e

\int \frac{2}{1 + 2 c o s x} d t =_{t a n (\frac{x}{2}) = t} \int \frac{2}{1 + 2 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 4 \int \frac{d t}{1 + t^{2} + 2 - 2 t^{2}}

= 4 \int \frac{d t}{3 - t^{2}} =_{t = \sqrt{3} u} 4 \int \frac{\sqrt{3} d u}{3 (1 - u^{2})} = \frac{2 \sqrt{3}}{4} \int \frac{2 d u}{1 - u^{2}} = \frac{\sqrt{3}}{2} l n ∣ \frac{1 + u}{1 - u} ∣ + c_{2}

= \frac{\sqrt{3}}{2} l n ∣ \frac{1 + \frac{t}{\sqrt{3}}}{1 - \frac{t}{\sqrt{3}}} ∣ + c_{2} = \frac{\sqrt{3}}{2} l n ∣ \frac{\sqrt{3} + t}{\sqrt{3} - t} ∣ + c_{2} f i n a l l y

A = l n ∣ t a n (\frac{x}{2} + \frac{π}{4}) ∣ - \frac{\sqrt{3}}{2} l n ∣ \frac{\sqrt{3} + t a n (\frac{x}{2})}{\sqrt{3} - t a n (\frac{x}{2})} ∣ + C .

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

\int \frac{d x}{c o s x + 2 {c o s}^{2} x}

\int \frac{d x}{c o s x (1 + 2 c o s x)}

\int \frac{1 + 2 c o s x - 2 c o s x}{c o s x (1 + 2 c o s x)} d x

\int \frac{d x}{c o s x} - \int \frac{2 d x}{2 (\frac{1}{2} + c o s x)}

\int s e c x d x - \int \frac{d x}{\frac{1}{2} + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}

\int s e c x d x - \int \frac{{s e c}^{2} \frac{x}{2}}{\frac{3}{2} - \frac{1}{2} {t a n}^{2} \frac{x}{2}} d x

\int s e c x d x - 2 \int \frac{{s e c}^{2} \frac{x}{2}}{3 - {t a n}^{2} \frac{x}{2}} d x

I_{1} - 2 I_{2}

I_{1} = \int s e c x d x

= l n ∣ s e x + t a n x ∣ + c_{1}

I_{2} = \int \frac{{s e c}^{2} \frac{x}{2}}{3 - {t a n}^{2} \frac{x}{2}} d x

t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x

I_{2} = \int \frac{2 d t}{{(\sqrt{3})}^{2} - t^{2}}

= 2 \times \frac{1}{2 \sqrt{3}} l n ∣ \frac{\sqrt{3} + t}{\sqrt{3} - t} ∣ + c_{2}

= \frac{1}{\sqrt{3}} l n ∣ \frac{\sqrt{3} + t a n \frac{x}{2}}{\sqrt{3} - t a n \frac{x}{2}} ∣ + c_{2}

Commented by maxmathsup by imad last updated on 05/Nov/18

t h a n k s s i r .

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

t h a n k y o u s i r \dots

Answered by ajfour last updated on 04/Nov/18

I = \int \frac{d x}{2 \cos^{2} x + \cos x}

l e t \tan \frac{x}{2} = t \Rightarrow d x = \frac{2 d t}{1 + t^{2}}

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

= \int \frac{2 (1 + t^{2}) d t}{2 {(1 - t^{2})}^{2} + 1 - t^{4}}

= 2 \int \frac{(1 + t^{2}) d t}{t^{4} - 4 t^{2} + 3} = 2 \int \frac{(\frac{1}{t^{2}} + 1) d t}{t^{2} + \frac{3}{t^{2}} - 4}

l e t a (1 - \frac{\sqrt{3}}{t^{2}}) + b (1 + \frac{\sqrt{3}}{t^{2}}) = \frac{1}{t^{2}} + 1

\Rightarrow a + b = 1 & b - a = \frac{1}{\sqrt{3}}

\Rightarrow a = \frac{1}{2} - \frac{1}{2 \sqrt{3}}; b = \frac{1}{2} + \frac{1}{2 \sqrt{3}}

I = (1 - \frac{1}{\sqrt{3}}) \int \frac{(1 - \frac{\sqrt{3}}{t^{2}}) d t}{{(t + \frac{\sqrt{3}}{t})}^{2} - 4 - 2 \sqrt{3}}

+ (1 + \frac{1}{\sqrt{3}}) \int \frac{(1 + \frac{\sqrt{3}}{t^{2}}) d t}{{(t - \frac{\sqrt{3}}{t})}^{2} - 4 + 2 \sqrt{3}}

I = (\frac{\sqrt{3} - 1}{\sqrt{3}}) \frac{1}{2 \sqrt{4 + 2 \sqrt{3}}} \ln ∣ \frac{t + \frac{\sqrt{3}}{t} - \sqrt{4 + 2 \sqrt{3}}}{t + \frac{\sqrt{3}}{t} + \sqrt{4 + 2 \sqrt{3}}} ∣

+ (\frac{\sqrt{3} + 1}{\sqrt{3}}) \frac{1}{2 \sqrt{4 - 2 \sqrt{3}}} \ln ∣ \frac{t + \frac{\sqrt{3}}{t} - \sqrt{4 - 2 \sqrt{3}}}{t + \frac{\sqrt{3}}{t} + \sqrt{4 - 2 \sqrt{3}}} ∣ + c .

Commented by maxmathsup by imad last updated on 04/Nov/18

t h a n k y o u s i r A j f o u r