pi-4-pi-2-4cot-x-1-4-cot-x-dx-

Question Number 117176 by bemath last updated on 10/Oct/20

\underset{π / 4}{\int^{π / 2}} \frac{4 \cot x + 1}{4 - \cot x} d x = ?

Commented by bemath last updated on 10/Oct/20

t h e o t h e r w a y

\frac{4 \cot x + 1}{4 - \cot x} = \cot (x - \cot^{- 1} (4))

\underset{π / 4}{\int^{π / 2}} \frac{4 \cot x + 1}{4 - \cot x} d x = [\ln {(\sin (x - \cot^{- 1} (4))]}_{π / 4}^{π / 2}

= \ln (\sin (\frac{π}{2} - \cot^{- 1} (4)) - \ln (\sin (\frac{π}{4} - \cot^{- 1} (4))

= \ln (\frac{\cos (\cot^{- 1} (4))}{\sin (\frac{π}{4} - \cot^{- 1} (4)})

= \ln (\frac{\frac{4}{\sqrt{17}}}{\frac{\sqrt{2}}{2} (\frac{4}{\sqrt{17}} - \frac{1}{\sqrt{17}})})

= \ln (\frac{4 \sqrt{2}}{3})

Answered by TANMAY PANACEA last updated on 10/Oct/20

\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{4 c o s x + s i n x}{4 s i n x - c o s x} d x

\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d (4 s i n x - c o s x)}{4 s i n x - c o s x}

∣ l n (4 s i n x - c o s x) ∣_{\frac{π}{4}}^{\frac{π}{2}}

= l n (4 - 0) - l n (4 \times \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})

l n (\frac{4 \sqrt{2}}{3})

Commented by bemath last updated on 10/Oct/20

t h a n k y o u s i r

Answered by AbduraufKodiriy last updated on 10/Oct/20

S o l u t i o n :

I = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{4 c o t (x) + 1}{4 - c o t (x)} d x = \int_{\frac{π}{4}}^{\frac{π}{2}} c o t (x - a r c c o t 4) d x =

= l n ∣ s i n (x - a r c c o t 4) ∣ ∣_{\frac{π}{4}}^{\frac{π}{2}} = l n (\frac{s i n (\frac{π}{2} - a r c c o t 4)}{s i n (\frac{π}{4} - a r c c o t 4)}) =

= l n (\frac{c o s (a r c c o s \frac{4}{\sqrt{17}})}{\frac{1}{\sqrt{2}} (c o s (a r c c o s \frac{4}{\sqrt{17}}) - s i n (a r c s i n \frac{1}{\sqrt{17}}))}) =

= l n (\frac{\frac{4}{\sqrt{17}}}{\frac{1}{\sqrt{2}} \cdot \frac{3}{\sqrt{17}}}) = l n \frac{4 \sqrt{2}}{3}

Commented by bemath last updated on 10/Oct/20

h a h a . . s a m e s i r

Commented by AbduraufKodiriy last updated on 10/Oct/20

Y e a h :)

Answered by Dwaipayan Shikari last updated on 10/Oct/20

\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{4 c o s x + s i n x}{4 s i n x - c o s x} d x

= {[l o g (4 s i n x - c o s x)]}_{\frac{π}{4}}^{\frac{π}{2}} = 2 l o g (2) - l o g (\frac{3}{\sqrt{2}})

= \frac{5}{2} l o g (2) - l o g (3)

Answered by mathmax by abdo last updated on 10/Oct/20

A = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{4 cotan (x) + 1}{4 - cotanx} dx \Rightarrow A = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\frac{4}{tanx} + 1}{4 - \frac{1}{tanx}} dx

= \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{4 + tanx}{4 tanx - 1} dx =_{tanx = t} \int_{1}^{\infty} \frac{4 + t}{4 t - 1} \frac{dt}{1 + t^{2}}

= \int_{1}^{\infty} \frac{t + 4}{(4 t - 1) (t^{2} + 1)} dt let decompose F (t) = \frac{t + 4}{(4 t - 1) (t^{2} + 1)}

F (t) = \frac{a}{4 t - 1} + \frac{bt + c}{t^{2} + 1}

a = (4 t - 1) F (t) ∣_{t = \frac{1}{4}} = \frac{\frac{1}{4} + 4}{\frac{1}{16} + 1} = \frac{17}{4} \times \frac{16}{17} = 4

\lim_{t \to + \infty} tF (t) = 0 = \frac{a}{4} + b \Rightarrow b = - 1

F (0) = - 4 = - a + c \Rightarrow c = a - 4 = 0 \Rightarrow F (t) = \frac{4}{4 t - 1} + \frac{- t}{t^{2} + 1} \Rightarrow

\int_{1}^{\infty} F (t) dt = \int_{1}^{\infty} (\frac{1}{t - \frac{1}{4}} - \frac{1}{2} \frac{2 t}{t^{2} + 1}) dt

= {[\ln ∣ \frac{t - \frac{1}{4}}{\sqrt{t^{2} + 1}} ∣]}_{1}^{\infty} = - \ln ∣ \frac{\frac{3}{4}}{\sqrt{2}} ∣ = - \ln ∣ \frac{3}{4 \sqrt{2}} ∣ = - \ln (3) + \ln (4 \sqrt{2})

= 2 \ln (2) + \frac{1}{2} \ln 2 - \ln (3) = \frac{5}{2} \ln (2) - \ln (3)