∫_0 ^( π/2) tan^3 xdx = ?


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Question Number 65950 by ajfour last updated on 06/Aug/19

$\int_{0}^{π / 2} \tan^{3} x d x = ?$
	Answered by Tupac Shakur last updated on 06/Aug/19
	$=∫_0 ^(π/2) ((sin^3 (x))/(cos^3 (x)))dx=∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx let u=cos(x)==>du=−sin(x)dx ∫_0 ^(π/2) ((sin(x){1−cos^2 (x)})/(cos^3 (x)))dx=∫_0 ^1 ((1−u^2 )/u^3 )du=lim_(t→0) ∫_t ^1 (1/u^3 )−(1/u^ )du=limt→0{−(3/2)+(1/(2t^2 ))+ln(t)}=+∞$
	$= \int_{0}^{\frac{π}{2}} \frac{{s i n}^{3} (x)}{{c o s}^{3} (x)} d x = \int_{0}^{\frac{π}{2}} \frac{s i n (x) {1 - {c o s}^{2} (x)}}{{c o s}^{3} (x)} d x$ $l e t u = c o s (x) ==> d u = - s i n (x) d x$ $\int_{0}^{\frac{π}{2}} \frac{s i n (x) {1 - {c o s}^{2} (x)}}{{c o s}^{3} (x)} d x = \int_{0}^{1} \frac{1 - u^{2}}{u^{3}} d u = l i \underset{t \to 0}{m} \int_{t}^{1} \frac{1}{u^{3}} - \frac{1}{u^{}} d u = l i m t \to 0 {- \frac{3}{2} + \frac{1}{2 t^{2}} + l n (t)} = + \infty$
	Answered by Tanmay chaudhury last updated on 06/Aug/19

	$\int {t a n}^{2} x . t a n x d x$ $\int ({s e c}^{2} x - 1) t a n x d x$ $\int t a n x . d (t a n x) - \int t a n x d x$ $\frac{{t a n}^{2} x}{2} - l n s e c x + c$ $w h e n w e p u t x = \frac{π}{2} e x p r e s s i o n \to \infty$ $s o s o m e d o u b t . . .$
	Answered by ajfour last updated on 06/Aug/19

	$\int_{0}^{π / 2} \frac{\tan^{3} x \sec^{2} x d x}{(1 + \tan^{2} x)}$ $l e t \tan x = t$ $\int_{0}^{\infty} \frac{t^{3} d t}{1 + t^{2}} = \int_{0}^{\infty} \frac{t (t^{2} + 1 - 1) d t}{1 + t^{2}}$ $= \int_{0}^{c \to \infty} [(t d t) - \frac{1}{2} (\frac{2 t d t}{1 + t^{2}})]$ $= \lim_{b \to \infty} [\frac{b}{2} - \frac{1}{2} \ln (1 + b)]$ $f (b) = \lim_{b \to \infty} \frac{\ln (1 + b)}{b} = \lim_{b \to \infty} \frac{1}{1 + b} \to 0$ $\Rightarrow I n t e g r a l \to \infty$ $(b u t a n s w e r i n b o o k R . S . A g a r w a l$ $f o r c l a s s X I I i s a f i n i t e o n e!)$
	Commented by MJS last updated on 06/Aug/19

	$post his answer please . . .$
	Answered by MJS last updated on 06/Aug/19

	$\int \tan^{n} x d x = \frac{1}{n - 1} \tan^{n - 1} x - \int \tan^{n - 2} x d x$ $\int \tan^{3} x d x = \frac{1}{2} \tan^{2} x - \int \tan x d x =$ $= \frac{1}{2} \tan^{2} x + \ln \cos x + C$ $\underset{0}{\int^{\frac{π}{2}}} \tan^{n} x d x = + \infty for n \in N^{★}$ ${what}^{'} s obvious because \tan^{n} x ⩾ 1 for \frac{π}{4} ⩽ x < \frac{π}{2}$ $and n \in N^{★}$
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