Question

Find the following integrals :-

(1). \: [tex\] \int \frac{dx}{
{x}^{2} - 16} \\ \\ (2). \: \int \frac{dx}{ \sqrt{2x - {x}^{2} } } [tex/]​

Answer 1

1. Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{x^2-16}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{(x-4)(x+4)}$

Multiplying and dividing the integrand by 4 - (- 4) = 8,

$\displaystyle\longrightarrow I=\dfrac{1}{8}\int\dfrac{8}{(x-4)(x+4)}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\int\dfrac{(x+4)-(x-4)}{(x-4)(x+4)}\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\int\left(\dfrac{1}{x-4}-\dfrac{1}{x+4}\right)\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{8}\big[\log|x-4|-\log|x+4|\big]+C$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{8}\log\left|\dfrac{x-4}{x+4}\right|+C}}$

2. Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{2x-x^2}}$

Pre-adding and subtracting (2/2)² = 1 in the denominator surd,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{1-1+2x-x^2}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{1-(x^2-2x+1)}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\sqrt{1^2-(x-1)^2}}$

We have,

• $\displaystyle\int\dfrac{dx}{\sqrt{c^2-(ax+b)^2}}=\dfrac{1}{a}\,\sin^{-1}\left(\dfrac{ax+b}{c}\right)+C$

Here,

• $a=1$

• $b=-1$

• $c=1$

Hence the integral becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{1}\,\sin^{-1}\left(\dfrac{x-1}{1}\right)+C$

$\displaystyle\longrightarrow\underline{\underline{I=\sin^{-1}\left(x-1\right)+C}}$

Answer 2

$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \rm :\implies\: \boxed{ \pink{ \bf \: \int \frac{dx}{</p><p>{x}^{2} - {a}^{2} }\: = \tt \: \dfrac{1}{2a} log(\dfrac{x - a}{x + a} ) \: + \: c}}$

$(2). \: \rm :\implies\: \boxed{ \pink{ \bf \: \int \: \dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } \: = \tt \: {sin}^{ - 1}\dfrac{x}{a} \: + \: c }}$

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$\large\underline\purple{\bold{Solution - }}$

$\bf \: (1). \: \int \dfrac{dx}{</p><p>{x}^{2} - 16}$

$\rm :\implies\:\int \dfrac{dx}{</p><p>{x}^{2} - {4}^{2} }$

$\rm :\implies\:\dfrac{1}{2 \times 4} \: log(\dfrac{x \: - \: 4}{x \: + \: 4} ) \: + \: c$

$\rm :\implies\:\dfrac{1}{8} \: log(\dfrac{x \: - \: 4}{x \: + \: 4} ) \: + \: c$

$\rm :\implies\: \boxed{ \green{ \bf \: \int \dfrac{dx}{</p><p>{x}^{2} - 16}\: = \tt \: \dfrac{1}{8} log(\dfrac{x - 4}{x + 4} ) \: + \: c}}$

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$\bf \: (2). \: \int \dfrac{dx}{ \sqrt{2x - {x}^{2} } }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{ -( {x}^{2} - 2x)} }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{ -( {x}^{2} - 2x + 1 - 1) } }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{ - \bigg( {(x - 1)}^{2} - 1 \bigg) } }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{1 - {(x - 1)}^{2} } }$

$\rm :\implies\: {sin}^{ - 1} \dfrac{x - 1}{1} + c$

$\rm :\implies\: {sin}^{ - 1} (x - 1) + c$

$\rm :\implies\: \boxed{ \blue{ \bf \: \: \int \frac{dx}{ \sqrt{2x - {x}^{2} } }\: = \tt \: {sin}^{ - 1}(x - 1) \: + \: c}}$