Question

evaluate integrate x(tan-1x)2 dx. ]01

(upper limit is 1 and lower limit is 0)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

note first I will solve integration then I will put limit

$\int \: x {( { \tan}^{ - 1} (x))}^{2}dx$

using formula

$\int \: (uv) dx= \\ u \int vdx - \int (\frac{du}{dx} \int vdx)dx$

now( using ILATE) put

$u = {({ \tan }^{ - 1} x)}^{2}$

and

$v = x$

you will found that

$\int \: x {( { \tan}^{ - 1} (x))}^{2}dx = \\ {( { \tan}^{ - 1} x)}^{2} \times \frac{ {x}^{2} }{2} - \\ \int \: \frac{ {x}^{2} }{2} \frac{ ({2 \tan }^{ - 1} x }{1 + {x}^{2} }$

let

I'=

$\frac{ {x}^{2} }{}( \frac{ { \tan }^{ - 1} x }{1 + {x}^{2} })$

now let

${ \tan}^{ - 1} x = t \\ then \: \: \: dx =( {1 + x}^{2} )dt$

on substituting value

I'=

$\int \: t( { \tan }^{2} t)\: dt = \\ (\int t ({ \sec }^{2} t) )- \int \: { {t} }{} \\ \\ = (t(\tan(t) - \int1 \times \tan(t) ) - \frac{ {t}^{2} }{2} \\ = (t(\tan(t) - log_{}( \sec(t) ) ) - \frac{ {t}^{2} }{2}$

Now

$log_{}( \sec(t) ) = log((1 + { \tan( { \tan}^{ - 1}x ) })^{ \frac{1}{2} } ) = \frac{1}{2} log(1 + x)$

therefore

I'=

$(t(\tan(t) - log_{}( 1 + x ) - \frac{ {t}^{2} }{2}$

now substituting these value in I

$\frac{({ { \tan}^{ - 1} x) }^{2} {x}^{2} }{2} - \\ ({ \tan}^{ - 1} x)x + \frac{1}{2} log(1 + x) + \\ \frac{ {({ \tan }^{ - 1} x)}^{2} }{2} \\ = \frac{( { { \tan }^{ - 1} x)}^{2} }{2} ( {x}^{2} + 1) + \\ x { \tan }^{ - 1}x + \\ \frac{1}{2} log(1 + x)$

now on putting limit

(*note=I am taking only upper limit because tan0=0 and log1=0)

SO

I=

$\frac{1}{2} \times {( \frac{ { \pi} }{4}}^{2} )( {1}^{2} + 1) - 1( \frac{ \pi}{4} ) + \frac{1}{2} log(2) \\ = \frac{ { \pi}^{2} }{16} - \frac{ \pi}{4} + \frac{1}{2} log(2)$

thanks for asking

if you have any problem regarding to this solution

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