Question

$\displaystyle \sf\int \: arccosec \:( x ) + x \: \: \: \: dx$
​

Answer 1

$\huge \red{\underline{\mathfrak \green{Solution-}}}$

Given integral is

$\begin{gathered}\rm \blue{ \displaystyle \sf \int arc \: cosx \: dx} \end{gathered}$

can be rewritten as

$\rm \blue{= \:\displaystyle \sf \int {cos}^{ - 1}x \: dx }$

can be further rewritten as

$\rm \blue{= \:\displaystyle \sf \int 1. \: {cos}^{ - 1}x \: dx}$

Using integration by parts, we get

$\begin{gathered}\rm \blue{= {cos}^{ - 1}x\displaystyle \sf \int 1 \: dx \: - \: \displaystyle \sf \int \bigg[\dfrac{d}{dx}{cos}^{ - 1}x\displaystyle \sf \int 1 \: dx \bigg]dx}\end{gathered} \\\begin{gathered}\rm \blue{= \:{cos}^{ - 1}x \: (x) - \displaystyle \sf \int \frac{ - 1}{ \sqrt{1 - {x}^{2} } } \: (x) \: dx}\end{gathered}$

can be further rewritten as

$\begin{gathered}\rm \blue{ = \:x \: {cos}^{ - 1}x \: - \displaystyle \sf \int \frac{ - x}{ \sqrt{1 - {x}^{2} } } \: dx} \end{gathered}$

Now, to evaluate this integral, we use method of Substitution.

So, Substitute

$\begin{gathered}\rm \blue{ \sqrt{1 - {x}^{2} } = y} \end{gathered}\\\begin{gathered}\rm \blue{1 - {x}^{2} = {y}^{2}} \end{gathered}\\\begin{gathered}\rm \blue{ \sqrt{1 - {x}^{2} } = y} \end{gathered}\\\begin{gathered}\rm \blue{1 - {x}^{2} = {y}^{2}} \end{gathered} \\\begin{gathered}\rm \blue{- 2x \: dx \: = \: 2y \: dy} \end{gathered} \\\begin{gathered}\rm \blue{- x \: dx \: = \: y \: dy}\end{gathered}$

So, on substituting these values, we get

$\begin{gathered}\rm \blue{\sqrt{1 - {x}^{2} } = y}\end{gathered}\\\begin{gathered}\rm \blue{1 - {x}^{2} = {y}^{2}} \end{gathered}\\\begin{gathered}\rm \blue{- 2x \: dx \: = \: 2y \: dy}\end{gathered} \\\begin{gathered}\rm \blue{ - x \: dx \: = \: y \: dy} \end{gathered} \\\begin{gathered}\rm \blue{= \:x \: {cos}^{ - 1}x \: - \: \displaystyle \sf \int \frac{y \: dy}{y}} \end{gathered} \\\begin{gathered}\rm \blue{ = \:x \: {cos}^{ - 1}x \: - \: \displaystyle \sf \int dy} \end{gathered}\\\begin{gathered}\rm \blue{= \:x \: {cos}^{ - 1}x \: - \: y \: + \: c}\end{gathered}\\\begin{gathered}\rm \blue{= \:x \: {cos}^{ - 1}x \: - \: \sqrt{1 - {x}^{2} } \: + \: c }\end{gathered}$

Hence,

$\begin{gathered} \red{\boxed{\sf \green{{ \: \: \rm \: \displaystyle \sf \int arc \: cosx= \:x \: {cos}^{ - 1}x \: - \: \sqrt{1 - {x}^{2} } \: + \: c \: \: }}}} \end{gathered}$

Formulae Used :-

Integration by parts :-

$\begin{gathered}\displaystyle \sf \blue{\int u. \: v \: dx \: = \: u\displaystyle \sf \int v \: dx \: - \: \displaystyle \sf \int \bigg[\dfrac{d}{dx} u \: \displaystyle \sf \int v \: dx\bigg]dx}\end{gathered}$

where u and v are chosen according to the word ILATE.

I : Inverse Trigonometric function

L : Logarithmic function

A : Arithmetic function

T : Trigonometric function

E : Exponential function

which alphabet comes first, is preferred to take as u and other as v.

$\begin{gathered}\rm \blue{ \displaystyle \rm \int {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c } \end{gathered} \\\begin{gathered}\rm \blue{ \dfrac{d}{dx}{cos}^{ - 1}x \: = \: \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } }} \end{gathered}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf \red{ f(x) }&\red{ \bf \displaystyle \int \rm \:f(x) \: dx}\\ \\ \red{\frac{\qquad \qquad}{}} & \red{\frac{\qquad \qquad}{}} \\ \sf\red{ k} & \sf\red{ kx + c }\\ \\ \sf\red{ sinx }& \sf\red{ - \: cosx+ c} \\ \\ \sf \red{cosx} & \sf \red{\: sinx + c}\\ \\ \sf\red{ {sec}^{2} x} & \sf\red{ tanx + c}\\ \\ \sf\red{ {cosec}^{2}x} & \sf \red{- cotx+ c} \\ \\ \sf\red{ secx \: tanx} & \sf\red{ secx + c}\\ \\ \sf \red{cosecx \: cotx}& \sf \red{- \: cosecx + c}\\ \\ \sf\red{ tanx }& \sf\red{ logsecx + c}\\ \\ \sf \red{\dfrac{1}{x} }& \sf\red{ logx+ c}\\ \\ \sf \red{{e}^{x} }& \sf\red{ {e}^{x} + c}\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle \sf\int \:( arccosec \:( x ) + x) \: dx$

can be rewritten as

$\rm \: = \displaystyle \sf\int \: {cosec}^{ - 1}x \: dx \: + \: \displaystyle \sf\int \:x \: dx$

can be further rewritten as

$\rm \: = \displaystyle \sf\int \: 1.{cosec}^{ - 1}x \: dx \: + \: \frac{ {x}^{2} }{2} + c \cdots \: (1)$

Let us consider

$\rm \: \displaystyle \sf\int \: 1.{cosec}^{ - 1}x \: dx \:$

Now, using integration by parts, we get

$\rm \: = \:{cosec}^{ - 1}x\displaystyle \sf\int \:1dx - \displaystyle \sf\int \:\bigg[\dfrac{d}{dx}{cosec}^{ - 1}x\displaystyle \sf\int \:1dx \bigg]dx$

$\rm \: = \:{cosec}^{ - 1}x \times x + \displaystyle \sf\int \:\bigg[\dfrac{1}{x \sqrt{ {x}^{2} - 1 } } \times x \bigg]dx$

$\rm \: = \:x \: {cosec}^{ - 1}x+ \displaystyle \sf\int \:\dfrac{1}{\sqrt{ {x}^{2} - 1 } } dx$

$\rm \: = \: x \:{cosec}^{ - 1} x+ log |x + \sqrt{ {x}^{2} - 1} \: |$

So, on substituting this result in equation (1), we get

$\rm \: = x \: {cosec}^{ - 1}x+ log |x + \sqrt{ {x}^{2} - 1} \: | + \: \frac{ {x}^{2} }{2} + c$

Hence,

$\rm \displaystyle \sf\int [arccosec \:( x ) + x]dx=x {cosec}^{ - 1} x+ log |x + \sqrt{ {x}^{2} - 1} \: | + \frac{ {x}^{2} }{2} + c$

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Formulae Used :-

$\boxed{ \rm{ \:\displaystyle \sf\int \: {x}^{n}dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

Integration by Parts

$\blue{\boxed{ \sf{ \:\displaystyle \sf\int \:uv \: dx = u\displaystyle \sf\int \:vdx - \displaystyle \sf\int \:\bigg[\dfrac{d}{dx}u\displaystyle \sf\int \:vdx \bigg]dx \: }}}$

where,

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , u and v are according to the word ILATE.

where,

• I - Inverse trigonometric functions

• L - Logarithmic functions

• A - Arithmetic function

• T - Trigonometric functions

• E - Exponential functions

The alphabet which comes first is taken as u and other as v.

$\rule{190pt}{2pt}$

Additional Information :-

$\red{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}}$

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$