anybody please help q...∫cos^(−1) (√x) dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 47364 by olj55336@awsoo.com last updated on 09/Nov/18

$a n y b o d y p l e a s e h e l p$ $q . . . \int \cos^{- 1} \sqrt{x} d x$
Commented by maxmathsup by imad last updated on 09/Nov/18

$c h a n g e m e n t \sqrt{x} = t g i v e I = \int a r c c o s (\sqrt{x}) = \int 2 t a r c o s (t) d t$ $=_{b y p a r t s} t^{2} a r c o s t + \int t^{2} \frac{d t}{\sqrt{1 - t^{2}}} b u t$ $\int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t = \int t . \frac{t}{\sqrt{1 - t^{2}}} d t =_{b y p a r t s} - t \sqrt{1 - t^{2}} - \int \sqrt{1 - t^{2}} d t a n d$ $\int \sqrt{1 - t^{2}} d t =_{t = s i n u} \int c o s u c o s u d u = \int \frac{1 + c o (2 u)}{2}$ $= \frac{u}{2} + \frac{1}{4} s i n (2 u) = \frac{u}{2} + \frac{1}{2} s i n u c o s u = \frac{1}{2} a r c s i n t + \frac{1}{2} t \sqrt{1 - t^{2}} \Rightarrow$ $\int a r c o s (\sqrt{x}) = x a r c o s (\sqrt{x}) - \sqrt{x} \sqrt{1 - x} - \frac{1}{2} a r c s i n (\sqrt{x}) - \frac{1}{2} \sqrt{x} \sqrt{1 - x} + C$ $= x a r c o s (\sqrt{x}) - \frac{3}{2} \sqrt{x} \sqrt{1 - x} - \frac{1}{2} a r c s i n (\sqrt{x}) + C .$
	Answered by Smail last updated on 09/Nov/18

	$A = \int {c o s}^{- 1} (\sqrt{x}) d x$ $b y p a r t s$ $u = {c o s}^{- 1} (\sqrt{x}) \Rightarrow u^{'} = - \frac{1}{2 \sqrt{x}} \times \frac{1}{\sqrt{1 - x}}$ $v^{'} = 1 \Rightarrow v = x$ $A = {x c o s}^{- 1} (\sqrt{x}) + \frac{1}{2} \int \sqrt{\frac{x}{1 - x}} d x$ $B = \int \sqrt{\frac{x}{1 - x}} d x$ $t = \sqrt{\frac{x}{1 - x}} \Rightarrow 2 t d t = \frac{1}{{(1 - x)}^{2}} d x$ $x = \frac{t^{2}}{t^{2} + 1}$ $B = \int t \times (2 t {(1 - \frac{t^{2}}{t^{2} + 1})}^{2} d t$ $= 2 \int \frac{t^{2}}{{(t^{2} + 1)}^{2}} d t$ $t = t a n u \Rightarrow d t = (1 + {t a n}^{2} u) d u$ $\frac{d t}{{(1 + t^{2})}^{2}} = \frac{d u}{1 + {t a n}^{2} u}$ $B = 2 \int {t a n}^{2} u \times {c o s}^{2} u d u = 2 \int {s i n}^{2} u d u$ $= \int (1 - c o s (2 u)) d u$ $= u - \frac{s i n (2 u)}{2} + c = u - s i n (u) c o s (u) + c$ $B = {t a n}^{- 1} (t) - \frac{t}{1 + t^{2}} + c = {t a n}^{- 1} (\sqrt{\frac{x}{1 - x}}) - \sqrt{x (1 - x)} + c$ $A = {x c o s}^{- 1} (\sqrt{x}) + \frac{1}{2} {t a n}^{- 1} (\sqrt{\frac{x}{1 - x}}) - \frac{1}{2} \sqrt{x (1 - x}) + C$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum