I= ∫_(0 ) ^1 x ln (1+x^2 ) dx ? I=∫ (√(sin x)) .cos^3 x dx ?


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Question Number 115464 by bemath last updated on 26/Sep/20

$I = \underset{0}{\int^{1}} x \ln (1 + x^{2}) d x ?$ $I = \int \sqrt{\sin x} . \cos^{3} x d x ?$
	Answered by malwaan last updated on 26/Sep/20
	$∫(√(sin x)) cos^3 x dx =∫(√(sin x)) cos^2 x cos x dx =∫(√(sin x)) (1−sin^2 x)d(sin x) =∫(√u)(1−u^2 )du {u=sin x} y=(√u)⇒y^2 =u⇒2ydy=du ∴ 2∫y(1−y^4 )ydy=2∫(y^2 −y^6 )dy = 2((y^3 /3) − (y^6 /6))+ C = 2(((((√u))^3 )/3) − ((((√u))^6 )/6) + C = 2(((u(√u))/3) − (u^3 /6)) + C = 2(((sinx(√(sinx)))/3) − ((sin^3 x)/6))+ C$
	$\int \sqrt{s i n x} {c o s}^{3} x d x$ $= \int \sqrt{s i n x} {c o s}^{2} x c o s x d x$ $= \int \sqrt{s i n x} (1 - {s i n}^{2} x) d (s i n x)$ $= \int \sqrt{u} (1 - u^{2}) d u {u = s i n x}$ $y = \sqrt{u} \Rightarrow y^{2} = u \Rightarrow 2 y d y = d u$ $∴ 2 \int y (1 - y^{4}) y d y = 2 \int (y^{2} - y^{6}) d y$ $= 2 (\frac{y^{3}}{3} - \frac{y^{6}}{6}) + C$ $= 2 (\frac{{(\sqrt{u})}^{3}}{3} - \frac{{(\sqrt{u})}^{6}}{6} + C$ $= 2 (\frac{u \sqrt{u}}{3} - \frac{u^{3}}{6}) + C$ $= 2 (\frac{s i n x \sqrt{s i n x}}{3} - \frac{{s i n}^{3} x}{6}) + C$
	Commented by bemath last updated on 26/Sep/20

	$g a v e k u d o s$
	Commented by malwaan last updated on 26/Sep/20

	$g a v e k u d o s$ $W h a t i s t h i s ? p l e a s e$
	Commented by bemath last updated on 26/Sep/20

	$g a v e k u d o s = g i v e c o m p l i m e n t s$
	Commented by bemath last updated on 26/Sep/20
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	Answered by malwaan last updated on 26/Sep/20

	$\int x l n (1 + x^{2}) d x$ $u = l n (1 + x^{2}) \Rightarrow d u = \frac{2 x}{1 + x^{2}} d x$ $d v = x d x \Rightarrow v = \frac{x^{2}}{2}$ $\int u d v = u v - \int v d u$ $∴ \int x l n (1 + x^{2}) d x =$ $\frac{x^{2}}{2} l n (1 + x^{2}) - \int \frac{x^{3}}{1 + x^{2}} d x$ $= \frac{x^{2}}{2} l n (1 + x^{2}) - \int [x - \frac{x}{1 + x^{2}}] d x$ $= \frac{x^{2}}{2} l n (1 + x^{2}) - \frac{x^{2}}{2} + \frac{1}{2} l n (1 + x^{2}) + C$ $∴_{0} \int^{1} x l n (1 + x^{2}) d x$ $= [\frac{1}{2} l n (2) - \frac{1}{2} + \frac{1}{2} l n (2)] - [0]$ $= l n (2) - \frac{1}{2}$
	Answered by Dwaipayan Shikari last updated on 26/Sep/20

	$\int_{0}^{1} xlog (1 + x^{2}) dx$ $= {[\log (1 + x^{2}) \frac{x^{2}}{2}]}_{0}^{1} - \int_{0}^{1} \frac{x^{2}}{2} . \frac{2 x}{1 + x^{2}}$ $= \frac{1}{2} \log (2) - \int_{0}^{1} x - \frac{x}{1 + x^{2}}$ $= \frac{1}{2} \log (2) - \frac{1}{2} + {[\frac{1}{2} \log (1 + x^{2})]}_{0}^{1}$ $= \log (2) - \frac{1}{2}$
	Answered by Ar Brandon last updated on 26/Sep/20

	$I = \int_{0}^{1} x \ln (1 + x^{2}) d x = \frac{1}{2} \int_{0}^{1} 2 x \ln (1 + x^{2}) d x$ $= \frac{1}{2} \int_{0}^{1} \ln (1 + x^{2}) d (x^{2}) = {[\frac{(1 + x^{2}) [\ln (1 + x^{2}) - 1]}{2}]}_{0}^{1}$ $= (\ln 2 - 1) + \frac{1}{2} = \ln 2 - \frac{1}{2}$
	Answered by mathmax by abdo last updated on 26/Sep/20

	$I = \int_{0}^{1} xln (1 + x^{2}) dx by parts we get$ $I = {[\frac{x^{2}}{2} \ln (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{x^{2}}{2} . \frac{2 x}{1 + x^{2}} dx = \frac{\ln (2)}{2} - \int_{0}^{1} \frac{x^{3}}{x^{2} + 1} dx but$ $\int_{0}^{1} \frac{x^{3}}{x^{2} + 1} dx = \int_{0}^{1} \frac{x (x^{2} + 1) - x}{x^{2} + 1} dx = \int_{0}^{1} xdx - \int_{0}^{1} \frac{xdx}{x^{2} + 1}$ $= \frac{1}{2} - {[\frac{1}{2} \ln (1 + x^{2})]}_{0}^{1} = \frac{1}{2} - \frac{1}{2} \ln (2) \Rightarrow I = \frac{\ln (2)}{2} - \frac{1}{2} + \frac{\ln (2)}{2}$ $= \ln (2) - \frac{1}{2}$
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