please-solve-0-pi-4-tan-9-x-dx-

Question Number 116196 by mnjuly1970 last updated on 01/Oct/20

p l e a s e s o l v e :

\int_{0}^{\frac{π}{4}} {t a n}^{9} (x) d x = ? ? ?

Answered by mindispower last updated on 01/Oct/20

l e t

u_{n} = \int_{0}^{\frac{π}{4}} {t g}^{n} (x) d x

U_{n + 2} + U_{n} = \int_{0}^{\frac{π}{4}} {t g}^{n} (x) d x + \int_{0}^{\frac{π}{4}} {t g}^{n + 2} (x) d x

= \int_{0}^{\frac{π}{4}} {t g}^{n} (x) (1 + {t g}^{2} (x)) d x = \int_{0}^{\frac{π}{4}} {t g}^{n} (x) d (t g (x))

= {[\frac{{t g}^{n + 1} (x)}{n + 1}]}_{0}^{\frac{π}{4}} = \frac{1}{n + 1}

\Rightarrow U_{n + 2} = \frac{1}{n + 1} - U_{n}

U_{0} = \frac{π}{4}, U_{1} = \int_{0}^{\frac{π}{4}} t g (x) d x = {[- l n (c o s (x))]}_{0}^{\frac{π}{4}}

= l n (\sqrt{2})

U_{3} = - U_{1} + \frac{1}{2}

U_{5} = - U_{3} + \frac{1}{4}

U_{7} = - U_{5} + \frac{1}{6}

U_{9} = - U_{7} + \frac{1}{8}

= \frac{1}{8} - \frac{1}{6} + \frac{1}{4} - \frac{1}{2} + l n (\sqrt{2})

= \frac{3 - 4 + 6 - 12}{24} + l n (\sqrt{2})

= \frac{- 7 + 12 l n (2)}{24}

Commented by mnjuly1970 last updated on 01/Oct/20

t h a n k y o u s i r . .

Answered by MJS_new last updated on 01/Oct/20

\int \tan^{9} x d x =

[t = \tan x \to d x = \frac{d t}{t^{2} + 1}]

= \int \frac{t^{9}}{t^{2} + 1} d t = \int (\frac{t}{t^{2} + 1} + t^{7} - t^{5} + t^{3} - t) d t =

= \frac{1}{2} \ln (t^{2} + 1) + \frac{t^{8}}{8} - \frac{t^{6}}{6} + \frac{t^{4}}{4} - \frac{t^{2}}{2}

\Rightarrow

answer is \frac{1}{2} \ln 2 - \frac{7}{24}

Commented by mnjuly1970 last updated on 01/Oct/20

g r a t e f u l . .

Answered by mathmax by abdo last updated on 01/Oct/20

let u_{n} = \int_{0}^{\frac{π}{4}} \tan^{2 n + 1} xdx \Rightarrow u_{n} = \int_{0}^{\frac{π}{4}} \tan^{2 n - 1} x (1 + \tan^{2} x - 1) dx

= \int_{0}^{\frac{π}{4}} \tan^{2 n - 1} (1 + \tan^{2} x) dx - u_{n - 1} we hsve

\int_{0}^{\frac{π}{4}} {(1 + \tan^{2} x)}^{2 n - 1} dx = {[\frac{1}{2 n} \tan^{2 n} x]}_{0}^{\frac{π}{4}} = \frac{1}{2 n} \Rightarrow u_{n} = \frac{1}{2 n} - u_{n - 1} \Rightarrow

u_{n} + u_{n - 1} = \frac{1}{2 n} \Rightarrow \sum_{k = 1}^{n} {(- 1)}^{k} (u_{k} + u_{k - 1}) = \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \Rightarrow

- (u_{1} + u_{0}) + (u_{2} + u_{1}) + \dots {(- 1)}^{n - 1} (u_{n - 1} + u_{n - 2}) + {(- 1)}^{n} (u_{n} + u_{n - 1})

= \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \Rightarrow - u_{0} + {(- 1)}^{n} u_{n} = \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \Rightarrow

{(- 1)}^{n} u_{n} = u_{0} + \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k}

u_{0} = \int_{0}^{\frac{π}{4}} \frac{sinx}{cosx} dx = {[- \ln ∣ cosx ∣]}_{0}^{\frac{π}{4}} = - \ln (\frac{1}{\sqrt{2}}) = \frac{1}{2} \ln (2) \Rightarrow

{(- 1)}^{n} u_{n} = \frac{1}{2} \ln (2) + \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k} \Rightarrow

u_{n} = \int_{0}^{\frac{π}{4}} \tan^{2 n + 1} x dx = {(- 1)}^{n} {\frac{\ln 2}{2} + \frac{1}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k}}

n = 4 \Rightarrow \int_{0}^{\frac{π}{4}} \tan^{9} xdx = \frac{\ln 2}{2} + \frac{1}{2} \sum_{k = 1}^{4} \frac{{(- 1)}^{k}}{k}

= \frac{\ln 2}{2} + \frac{1}{2} {- 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4}} = \frac{\ln 2}{2} + \frac{1}{2} {- \frac{4}{3} + \frac{3}{4}}

= \frac{\ln 2}{2} + \frac{1}{2} (\frac{- 7}{12}) = \frac{\ln 2}{2} - \frac{7}{24}

Answered by Dwaipayan Shikari last updated on 01/Oct/20

\int_{0}^{\frac{π}{4}} \frac{\sin^{9} x}{\cos^{9} x} dx = - \int_{0}^{\frac{π}{4}} \frac{\sin^{8} x (- sinx)}{\cos^{9} x} dx

- \int_{1}^{\frac{1}{\sqrt{2}}} \frac{{(1 - t^{2})}^{4}}{t^{9}} dt = \int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{t^{9}} - \frac{4}{t^{7}} + \frac{6}{t^{5}} - \frac{4}{t^{3}} + \frac{1}{t}

= {[- \frac{1}{{8 t}^{8}} + \frac{2}{{3 t}^{6}} - \frac{3}{{2 t}^{4}} + \frac{2}{t^{2}} + \log (t)]}_{\frac{1}{\sqrt{2}}}^{1} = \log (\sqrt{2}) - \frac{7}{24}

Answered by maths mind last updated on 01/Oct/20

\int_{0}^{\frac{π}{4}} {t g}^{a} (x) d x = f (a), a \in C R e (a) ⩾ 0

t g (x) = t \Rightarrow f (a) = \int_{0}^{1} \frac{t^{a}}{1 + t^{2}} d t = \int_{0}^{1} \sum_{m ⩾ 0} {(- t^{2})}^{m} t^{a} d t

= \sum_{m ⩾ 0} {(- 1)}^{m} \int_{0}^{1} t^{2 m + a} d t

= \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{2 m + a + 1} = \sum_{m ⩾ 0} (\frac{1}{4 m + a + 1} - \frac{1}{4 m + a + 3})

= \frac{1}{4} \sum_{m ⩾ 0} (\frac{\frac{a + 3}{4} - \frac{a + 1}{4}}{(m + \frac{a + 1}{4}) (m + \frac{a + 2}{4})}) = \frac{2}{16} . \frac{Ψ (\frac{a + 3}{4}) - Ψ (\frac{a + 1}{4})}{\frac{a + 3}{4} - \frac{a + 1}{4}}

= \frac{Ψ (\frac{a + 3}{4}) - Ψ (\frac{a + 1}{4})}{4} = f (a)

f (9) = \frac{Ψ (\frac{12}{4}) - Ψ (\frac{10}{4})}{4}

Ψ (z + 1) = Ψ (z) + \frac{1}{z}

Ψ (3) = - γ + 1 + \frac{1}{2} = \frac{3}{2} - γ

Ψ (\frac{10}{4}) = Ψ (\frac{5}{2}) = Ψ (\frac{3}{2}) + \frac{2}{3} = \frac{2}{3} + 2 + Ψ (\frac{1}{2}), Ψ (\frac{1}{2}) = - 2 l n (2) - γ

Ψ (3) - Ψ (\frac{5}{2}) = \frac{3}{2} - γ - \frac{2}{3} - 2 - Ψ (\frac{1}{2}) = \frac{3}{2} - γ - \frac{8}{3} + 2 l n (2) + γ

= \frac{- 7}{6} + 2 l n (2)

f (9) = \frac{- 7 + 12 l n (2)}{24} = - \frac{7}{24} + \frac{l n (2)}{2}

Commented by mnjuly1970 last updated on 02/Oct/20

v e r y n i c e . . t h a n k y o u s o

m u c h \dots .

Answered by 1549442205PVT last updated on 02/Oct/20

I_{9} = \int \tan^{9} xdx = \int \tan^{7} x (1 + \tan^{2} x) dx

- \int \tan^{7} xdx = \int \tan^{7} xd (tanx) - I_{7}

= \frac{\tan^{8} x}{8} - I_{7} = \frac{\tan^{8} x}{8} - \frac{\tan^{6} x}{6} + \frac{\tan^{4} x}{4}

- \frac{\tan^{2} x}{2} + I_{1} = \frac{\tan^{8} x}{8} - \frac{\tan^{6} x}{6} + \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} - lncosx

Hence, \int_{0}^{\frac{π}{4}} {t a n}^{9} (x) d x = I_{9} ∣_{0}^{π / 4}

= \frac{1}{8} - \frac{1}{6} + \frac{1}{4} - \frac{1}{2} - \ln \frac{1}{\sqrt{2}} = \frac{- 7}{24} + \frac{1}{2} \ln 2

Commented by mnjuly1970 last updated on 02/Oct/20

t h a n k s a l o t

Commented by 1549442205PVT last updated on 07/Oct/20

You are welcome .