Question

integrate the following

Answer 1

Step-by-step explanation:

$\huge\mathfrak\green{\bold{\underline{☘{ ℘ɧεŋσɱεŋศɭ}☘}}}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve and verify the equation

$\frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )$

$\huge\tt\underline\blue{Answer }$

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$⟹</p><p> \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{3} )$

$⟹</p><p> \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - ( \frac{2x + 3}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - \frac{2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{6x - 2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{4x - 3}{6}$

cancelling 6( R.H.S) By 3 From L.H.S

$⟹ \frac{x - 12}{1} = \frac{4x - 3}{2} </p><p>$

$⟹</p><p>2(x - 12) = 4x - 3$

$⟹</p><p>2x - 24 = 4x - 3$

$⟹</p><p> - 24 + 3 = 4x - 2x$

$⟹</p><p> - 21 = 2x$

$⟹</p><p>x = - \frac{21}{2}$

CHECK:-

$⟹ \frac{ - \frac{21}{2} }{3} - 4 = - \frac{21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} )</p><p>$

$⟹</p><p> - \frac{21}{6} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{21}{6} )$

$⟹</p><p> - \frac{7}{2} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} )$

$⟹</p><p> \frac{ - 7 - 8}{2} = - \frac{21}{2} - ( - \frac{6}{2} )$

$⟹ - \frac{15}{2} = - \frac{21}{2} - ( - 3) </p><p>$

$⟹</p><p> - \frac{15}{2} = - \frac{21}{2} + 3$

$⟹</p><p> - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}$

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

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Answer 2

Given Integrand,

$\displaystyle \sf y = \int {sin}^{ - 1} \bigg( \dfrac{2x}{1 + {x}^{2} } \bigg)dx$

Let x = tan(u).

Differentiating both sides w.r.t u,

$\longrightarrow \sf \: \dfrac{dx}{du} = {sec}^{2} u \\ \\ \longrightarrow \sf \: \: dx = {sec}^{2} u \: du$

Now,

$\implies \: \sf y =\displaystyle \int \sf {sin}^{ - 1} \bigg( \dfrac{2tan \: u}{1 + {tan}^{2}u } \bigg)dx \\ \\ \implies \: \sf y =\displaystyle \sf \int {sin}^{ - 1} \bigg( \dfrac{2tan \: u}{1 + {tan}^{2}u } \bigg) \: sec {}^{2}u du \\ \\ \implies \displaystyle \sf y = \int {sin}^{ - 1} \bigg( sin2u \bigg) \: sec {}^{2}u du \\ \\ \implies \displaystyle \sf y = 2\int u \: sec {}^{2}u du$

Integrating by parts,

$\implies \displaystyle \sf \: y = 2u \int {sec}^{2} u \: du - 2 \int \bigg \{ \dfrac{du}{du} \int \sec {}^{2}u \: du \bigg \} \: du$

We know that,

$\star \: \displaystyle \boxed{ \boxed{ \sf \int sec {}^{2} a \: da = tan \: a + c}}$

Now,

$\implies \displaystyle \sf \: y = 2u \: tan(u) - 2 \int tan \: u \: du$

Also,

$\star \: \displaystyle \boxed{ \boxed{ \sf \int tan \: a \: da = log(sec \: a) + c}}$

Thus,

$\implies \displaystyle \sf \: y = 2u \: tan(u) - 2 log |sec(u)| + c$

Substituting u = arctan(x),

$\implies \boxed{ \boxed{ \displaystyle \sf \: y = 2x \: tan {}^{ - 1} (x) - 2 log |sec(tan {}^{ - 1}x )| + c}}$