Evaluate a. ∫_1 ^2 (lnx)^2 dx b. ∫_0 ^(π/6) sin^2 x cos^3 xdx


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Question Number 66016 by Rio Michael last updated on 07/Aug/19

$E v a l u a t e$ $a . \int_{1}^{2} {(l n x)}^{2} d x$ $b . \int_{0}^{\frac{π}{6}} {s i n}^{2} x {c o s}^{3} x d x$
Commented by Prithwish sen last updated on 07/Aug/19

$b . \int_{0}^{\frac{π}{6}} \sin^{2} x (1 - \sin^{2} x) cosxdx$ $now put sinx = u \Rightarrow cosx dx = du$ $\int_{0}^{\frac{1}{2}} u^{2} (1 - u^{2}) du = \int_{0}^{\frac{1}{2}} (u^{2} - u^{4}) du = {[\frac{1}{3} u^{3} - \frac{1}{5} u^{5}]}_{0}^{\frac{1}{2}}$ $= \frac{1}{24} - \frac{1}{160} = \frac{17}{480} please check .$
Commented by Prithwish sen last updated on 07/Aug/19

$a . lnx = t \Rightarrow \frac{dx}{x} = dt \Rightarrow dx = e^{t} dt$ $\int_{0}^{\ln 2} t^{2} e^{t} dt = {[t^{2} e^{t} - {2 te}^{t} + {2 e}^{t}]}_{0}^{\ln 2} = 2 {(\ln 2)}^{2} - 4 \ln 2 + 2$ $please check .$
Commented by kaivan.ahmadi last updated on 07/Aug/19

$\int (1 - {c o s}^{2} x) {c o s}^{3} x d x = \int ({c o s}^{3} x d x - {c o s}^{5} x) d x =$ $\int c o s x (1 - {s i n}^{2} x - {(1 - {s i n}^{2} x)}^{2}) d x =$ $\int c o s x (1 - {s i n}^{2} x - 1 + 2 {s i n}^{2} x - {s i n}^{4} x) d x =$ $\int c o s x ({s i n}^{2} x - {s i n}^{4} x) d x$ $s e t u = s i n x \Rightarrow d u = c o s x d x$ $\int (u^{2} - u^{4}) d u = \frac{u^{3}}{3} - \frac{u^{5}}{5} = \frac{{s i n}^{3} x}{3} - \frac{{s i n}^{5} x}{5}$
Commented by mathmax by abdo last updated on 07/Aug/19
$a) changement lnx =t give ∫_1 ^2 (lnx)^2 dx =∫_0 ^(ln(2)) t^2 e^t dt by parts ∫_0 ^(ln(2)) t^2 e^t dt =[t^2 e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) 2t e^t dt =2(ln2)^2 −2 { [te^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) e^t dt} =2(ln(2))^2 −2{2ln(2)−(2−1)} =2(ln(2))^2 −4ln(2) +2$
$a) c h a n g e m e n t l n x = t g i v e \int_{1}^{2} {(l n x)}^{2} d x$ $= \int_{0}^{l n (2)} t^{2} e^{t} d t b y p a r t s$ $\int_{0}^{l n (2)} t^{2} e^{t} d t = {[t^{2} e^{t}]}_{0}^{l n (2)} - \int_{0}^{l n (2)} 2 t e^{t} d t$ $= 2 {(l n 2)}^{2} - 2 {{[{t e}^{t}]}_{0}^{l n (2)} - \int_{0}^{l n (2)} e^{t} d t}$ $= 2 {(l n (2))}^{2} - 2 {2 l n (2) - (2 - 1)}$ $= 2 {(l n (2))}^{2} - 4 l n (2) + 2$
Commented by mathmax by abdo last updated on 07/Aug/19
$b) let I =∫_0 ^(π/6) sin^2 x cos^3 x ⇒I =∫_0 ^(π/6) (sinxcosx)^2 cosxdx =(1/4)∫_0 ^(π/6) sin^2 2x cosx dx =(1/4) ∫_0 ^(π/6) ((1−cos(4x))/2) cosx dx =(1/8) ∫_0 ^(π/6) (cosx−cos(4x)cosx)dx =(1/8) ∫_0 ^(π/6) cosxdx −(1/8) ∫_0 ^(π/6) cos(4x)cosxdx =(1/8).(1/2) −(1/(16)) ∫_0 ^(π/6) (cos(5x)+cos3x)dx =(1/(16))−(1/(16))[(1/5)sin(5x)+(1/3)sin(3x)]_0 ^(π/6) =(1/(16))−(1/(16)){(1/(10)) +(1/3)} =(1/(16))−(1/(160))−(1/(48)) =....$
$b) l e t I = \int_{0}^{\frac{π}{6}} {s i n}^{2} x {c o s}^{3} x \Rightarrow I = \int_{0}^{\frac{π}{6}} {(s i n x c o s x)}^{2} c o s x d x$ $= \frac{1}{4} \int_{0}^{\frac{π}{6}} {s i n}^{2} 2 x c o s x d x = \frac{1}{4} \int_{0}^{\frac{π}{6}} \frac{1 - c o s (4 x)}{2} c o s x d x$ $= \frac{1}{8} \int_{0}^{\frac{π}{6}} (c o s x - c o s (4 x) c o s x) d x$ $= \frac{1}{8} \int_{0}^{\frac{π}{6}} c o s x d x - \frac{1}{8} \int_{0}^{\frac{π}{6}} c o s (4 x) c o s x d x$ $= \frac{1}{8} . \frac{1}{2} - \frac{1}{16} \int_{0}^{\frac{π}{6}} (c o s (5 x) + c o s 3 x) d x$ $= \frac{1}{16} - \frac{1}{16} {[\frac{1}{5} s i n (5 x) + \frac{1}{3} s i n (3 x)]}_{0}^{\frac{π}{6}}$ $= \frac{1}{16} - \frac{1}{16} {\frac{1}{10} + \frac{1}{3}} = \frac{1}{16} - \frac{1}{160} - \frac{1}{48} = . . . .$
Commented by Rio Michael last updated on 07/Aug/19

$T h a n k s s i r s$
	Answered by Tanmay chaudhury last updated on 07/Aug/19

	$b . \int_{0}^{\frac{π}{6}} {s i n}^{2} x \times (1 - {s i n}^{2} x) d (s i n x)$ $∣ \frac{{s i n}^{3} x}{3} - \frac{{s i n}^{5} x}{5} ∣_{0}^{\frac{π}{6}}$ $= \frac{{(\frac{1}{2})}^{3}}{3} - \frac{{(\frac{1}{2})}^{5}}{5}$ $\frac{1}{8} (\frac{1}{3} - \frac{1}{20})$ $= \frac{17}{480}$
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