Let-I-n-0-pi-4-tan-n-x-dx-n-gt-1-and-n-N-then-

Question Number 53696 by gunawan last updated on 25/Jan/19

Let I_{n} = \underset{0}{\int^{π / 4}} \tan^{n} x d x, (n > 1 and n \in N), then

Commented by Abdo msup. last updated on 25/Jan/19

s i r t h i s i n t e g r a l i s s o l v e d s e e t h e p l a t f o r m \dots

Commented by gunawan last updated on 25/Jan/19

thank you Sir

I {don}^{'} t know this integral is solved

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

J_{n} = \int {t a n}^{n - 2} {x t a n}^{2} x d x

j_{n} . = \int {t a n}^{n - 2} x ({s e c}^{2} x - 1) d x

= \int {t a n}^{n - 2} x d (t a n x) - \int {t a n}^{n - 2} x d x

j_{n} = \frac{({t a n}^{n - 2 + 1} x)}{n - 2 + 1} - j_{n - 2}

s o

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

I_{n} = \frac{1}{n - 1} - I_{n - 2} \leftarrow r e d u c t i o n f o r m u l a

I_{n - 2} = \frac{1}{n - 3} - I_{n - 4}

I_{n - 4} = \frac{1}{n - 5} - I_{n - 6}

\dots . .

\dots . .

I_{2} = \frac{1}{2 - 1} - I_{0}

n o w I_{0} = \int_{0}^{\frac{π}{4}} {(t a n x)}^{0} d x =∣ x ∣_{0}^{\frac{π}{4}} = \frac{π}{4}

n o w I_{2} . = \frac{1}{2 - 1} - \frac{π}{4} . = \frac{1}{1} - \frac{π}{4}

Commented by peter frank last updated on 25/Jan/19

n i c e w o r k s i r

Answered by peter frank last updated on 25/Jan/19

i t s r e d u c t i o n f o r m u l a r

I_{n} = \int {t a n}^{n} x d x

I_{n} = \int {t a n}^{n} x \tan^{2} x d x

I_{n} = \int {t a n}^{n} x \sec^{2} x d x - \int \tan^{n} x d x

b y p a r t

v = \tan x

\frac{d u}{d x} = \sec^{2} x

I_{n} = \frac{v^{n - 1} x}{n - 1} - I_{n - 2}

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

n ⩾ 2