Question

$\displaystyle \sf \int \arccos \: x \: dx$
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Answer 1

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

Given integral is

$\rm \: \displaystyle \sf \int arc \: cosx \: dx$

can be rewritten as

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

can be further rewritten as

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

Using integration by parts, we get

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

$\rm \: = \:{cos}^{ - 1}x \: (x) - \displaystyle \sf \int \frac{ - 1}{ \sqrt{1 - {x}^{2} } } \: (x) \: dx$

can be further rewritten as

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

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

So, Substitute

$\rm \: \sqrt{1 - {x}^{2} } = y$

$\rm \: 1 - {x}^{2} = {y}^{2}$

$\rm \: - 2x \: dx \: = \: 2y \: dy$

$\rm \: - x \: dx \: = \: y \: dy$

So, on substituting these values, we get

$\rm \: = \:x \: {cos}^{ - 1}x \: - \: \displaystyle \sf \int \frac{y \: dy}{y}$

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

$\rm \: = \:x \: {cos}^{ - 1}x \: - \: y \: + \: c$

$\rm \: = \:x \: {cos}^{ - 1}x \: - \: \sqrt{1 - {x}^{2} } \: + \: c$

Hence,

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

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Formulae Used :-

Integration by parts :-

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

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.

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

$\rm \: \dfrac{d}{dx}{cos}^{ - 1}x \: = \: \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } }$

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

Additional Information :-

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

Answer 2

Step-by-step explanation:

simple way.

Put $\tt\cos ^{-1} x=t$ so that $\tt x=\cos t$ and $\tt d x=-\sin t d t$

$\color{magenta}\begin{aligned} \tt \therefore \quad \int \cos ^{-1} x d x & \tt=-\int t \sin t d t \\ \\ & \tt=-\left[t \cdot(-\cos t)-\int 1 \cdot(-\cos t) d t\right] \\& \qquad\qquad \qquad\qquad \text{ [integrating \: \: by \: \: parts]} \\ \\ & \tt=t \cos t-\int \cos t d t\\ \\ & \tt=t \cos t-\sin t+C \\ \\ & \tt=x \cos ^{-1} x-\sqrt{1-x^{2}}+C \\ & \qquad \qquad\qquad \tt \left[\because \cos t=x \Rightarrow \sin t=\sqrt{1-x^{2}}\right] . \end{aligned}$