Evaluate-0-pi-2-1-1-cos-cos-x-dx-

Question Number 83164 by niroj last updated on 28/Feb/20

Evaluate :

\int_{0}^{\frac{π}{2}} \frac{1}{1 + \cos α \cos x} dx

Commented by mathmax by abdo last updated on 28/Feb/20

l e t I = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + c o s α c o s x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int_{0}^{1} \frac{1}{1 + c o s α \times \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{0}^{1} \frac{d t}{1 + t^{2} + c o s α (1 - t^{2})}

= 2 \int_{0}^{1} \frac{d t}{(1 - c o s α) t^{2} + 1 + c o s α} = \frac{2}{1 - c o s α} \int_{0}^{1} \frac{d t}{t^{2} + \frac{1 + c o s α}{1 - c o s α}}

= \frac{2}{1 - c o s α} \int_{0}^{1} \frac{d t}{t^{2} + {t a n}^{2} (\frac{α}{2})} =_{t =∣ t a n (\frac{α}{2}) ∣ u} \frac{2}{1 - c o s α} \int_{0}^{∣ c o t a n (\frac{α}{2}) ∣} \frac{∣ t a n (\frac{α}{2}) ∣ d u}{{t a n}^{2} (\frac{α}{2}) (1 + u^{2})}

= \frac{2}{(1 - c o s α) ∣ t a n (\frac{α}{2}) ∣} {[a r c t a n u]}_{0}^{∣ \frac{1}{t a n (\frac{α}{2})} ∣}

I = \frac{2}{(1 - c o s α) ∣ t a n (\frac{α}{2}) ∣} a r c t a n (∣ \frac{1}{t a n (\frac{α}{2})} ∣)

c a s e 1 t a n (\frac{α}{2}) > 0 \Rightarrow I = \frac{2 c o s (\frac{α}{2})}{2 {s i n}^{2} (\frac{α}{2}) \times s i n (\frac{α}{2})} (\frac{π}{2} - \frac{α}{2})

= \frac{c o s (\frac{α}{2})}{{s i n}^{3} (\frac{α}{2})} \times (\frac{π - α}{2})

c a s e 2 t a n (\frac{α}{2}) < 0 \Rightarrow I = - \frac{c o s (\frac{α}{2})}{{s i n}^{3} (\frac{α}{2})} \times (- a r c t a n (\frac{1}{t a n (\frac{α}{2})})

= \frac{c o s (\frac{α}{2})}{{s i n}^{3} (\frac{α}{2})} (- \frac{π}{2} - \frac{α}{2}) = - \frac{c o s (\frac{α}{2})}{{s i n}^{3} (\frac{α}{2})} (\frac{π + α}{2})

Commented by niroj last updated on 28/Feb/20

thank you both of you .

Answered by mr W last updated on 28/Feb/20

\frac{1}{1 + \cos α \cos x}

= \frac{1}{\cos α} \times \frac{1}{\frac{1}{\cos α} + \cos x}

l e t a = \frac{1}{\cos α} ⩾ 1 o r ⩽ - 1

\int_{0}^{\frac{π}{2}} \frac{1}{1 + \cos α \cos x} dx

= a \int_{0}^{\frac{π}{2}} \frac{d x}{a + \cos x}

= \frac{2 a}{\sqrt{a^{2} - 1}} {[\tan^{- 1} (\frac{\sqrt{a - 1}}{\sqrt{a + 1}} \tan \frac{x}{2})]}_{0}^{\frac{π}{2}}

= \frac{2 a}{\sqrt{a^{2} - 1}} \times \tan^{- 1} (\frac{\sqrt{a - 1}}{\sqrt{a + 1}})

= \frac{2}{\sqrt{1 - \frac{1}{a^{2}}}} \times \tan^{- 1} (\frac{\sqrt{1 - \frac{1}{a}}}{\sqrt{1 + \frac{1}{a}}})

= \frac{2}{\sqrt{1 - \cos^{2} α}} \times \tan^{- 1} (\frac{\sqrt{1 - \cos α}}{\sqrt{1 + \cos α}})

= \frac{2}{\sin α} \times \tan^{- 1} (\frac{\sqrt{1 - 1 + 2 \sin^{2} \frac{α}{2}}}{\sqrt{1 + 2 \cos^{2} \frac{α}{2} - 1}})

= \frac{2}{\sin α} \times \tan^{- 1} (\tan \frac{α}{2})

= \frac{2}{\sin α} \times \frac{α}{2}

= \frac{α}{\sin α}

Answered by TANMAY PANACEA last updated on 28/Feb/20

\int_{0}^{\frac{π}{2}} \frac{d x}{c o s α (\frac{1}{c o s α} + c o s x)} [\frac{1}{c o s α} = a]

\frac{1}{c o s α} \int_{0}^{\frac{π}{2}} \frac{d x}{a + c o s x}

\frac{1}{c o s α} \int_{0}^{\frac{π}{2}} \frac{d x}{a + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}

\frac{1}{c o s α} \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2}}{(a + 1) + {t a n}^{2} \frac{x}{2} \times (a - 1)} d x

\frac{1}{c o s α} \times \frac{1}{a - 1} \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2}}{\frac{a + 1}{a - 1} + {t a n}^{2} \frac{x}{2}} d x

\frac{1}{c o s α} \times \frac{1}{\frac{1}{c o s α} - 1} \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2}}{\frac{1 + c o s α}{1 - c o s α} + {t a n}^{2} \frac{x}{2}} d x

\frac{1}{1 - c o s α} \int_{0}^{\frac{π}{2}} \frac{d (t a n \frac{x}{2}) \times 2}{{c o t}^{2} \frac{α}{2} + {t a n}^{2} \frac{x}{2}}

\frac{2}{2 {s i n}^{2} \frac{α}{2}} \times \frac{1}{c o t \frac{α}{2}} \times ∣ {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{c o t \frac{α}{2}}) ∣_{0}^{\frac{π}{2}}

\frac{2}{s i n α} \times {t a n}^{- 1} (\frac{1}{c o t \frac{α}{2}}) = \frac{2}{s i n α} \times \frac{α}{2} = \frac{α}{s i n α}

p l s c h e c k \dots

Commented by niroj last updated on 28/Feb/20

great dear \dots

Commented by TANMAY PANACEA last updated on 28/Feb/20

t h a n k y o u s i r