0-pi-4-tan-n-x-dx-

Question Number 123552 by Lordose last updated on 26/Nov/20

\int_{0}^{\frac{π}{4}} \tan^{n} (x) dx

Answered by Adeleke last updated on 26/Nov/20

Answered by TANMAY PANACEA last updated on 26/Nov/20

I_{n} = \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} x ({s e c}^{2} x - 1) d x

= \int_{0}^{\frac{π}{4}} {t a n}^{n - 2} x \times d (t a n x) - I_{n - 2}

=∣ \frac{{t a n}^{n - 1} x}{n - 1} ∣_{0}^{\frac{π}{4}} - I_{n - 2}

= \frac{1}{n - 1} - I_{n - 2}

I_{n} = \frac{1}{n - 1} - I_{n - 2}

Commented by peter frank last updated on 26/Nov/20

thank you

Answered by Dwaipayan Shikari last updated on 26/Nov/20

I f i t w a s \int_{0}^{\frac{π}{2}} {t a n}^{n} x d x

t h e n w e c a n g e t a c l o s e d f o r m

\int_{0}^{\frac{π}{2}} {s i n}^{n} x {c o s}^{- n} x d x

= \frac{1}{2} \int_{0}^{1} t^{\frac{n}{2} - \frac{1}{2}} {(1 - t)}^{\frac{- n}{2} - \frac{1}{2}} d t {s i n}^{2} x = t

= \frac{1}{2} \frac{Γ (\frac{n}{2} + \frac{1}{2}) Γ (\frac{1}{2} - \frac{n}{2})}{Γ (1)} = \frac{π}{2 s i n (\frac{n π}{2} + \frac{π}{2})}

\int_{0}^{\frac{π}{4}} {t a n}^{p} x d x = \int_{0}^{1} \frac{t^{p}}{t^{2} + 1} d t

= \int_{0}^{1} t^{p} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} t^{2 n} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} t^{2 n + p} d t = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{2 n + p + 1}

Answered by mnjuly1970 last updated on 26/Nov/20

Ω \overset{t a n (x) = y}{=} \int_{0}^{1} \frac{y^{n} d y}{1 + y^{2}} = \int_{0}^{1} \frac{y^{n} - y^{n + 2}}{1 - y^{4}} d y

= \frac{1}{4} \int_{0}^{1} \frac{t^{\frac{n - 3}{4}} - t^{\frac{n - 1}{4}}}{1 - t} d t

= \frac{1}{4} (H_{\frac{n - 1}{4}} - H_{\frac{n - 3}{4}})

= \frac{1}{4} (ψ (\frac{n}{4} + \frac{3}{4}) - ψ (\frac{n}{4} + \frac{1}{4})) ✓

Commented by Dwaipayan Shikari last updated on 26/Nov/20

\int_{0}^{\frac{π}{4}} {t a n}^{4} x d x = \frac{1}{4} (ψ (1 + \frac{3}{4}) - ψ (1 + \frac{1}{4}))

= \frac{1}{4} (ψ (\frac{3}{4}) + \frac{4}{3} - ψ (\frac{1}{4}) - 4)

= \frac{1}{4} (π c o t (\frac{π}{4}) - \frac{8}{3}) = \frac{π}{4} - \frac{2}{3}

Commented by Lordose last updated on 02/Dec/20

Exactly

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