Question

$\tt\pink{Evaluate\displaystyle \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right)\tt d x}$
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Answer 1

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

Given integral is

$\rm \: \displaystyle \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right)d x$

We know that

$\boxed{\tt{ \: {cot}^{ - 1}x = {tan}^{ - 1} \frac{1}{x} \: }}$

So, using this, we get

$\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1}{1 - x + {x}^{2} } \bigg] d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1 - x + x}{1 - x + {x}^{2} } \bigg] d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{(1 - x) + x}{1 - x(1 - x)} \bigg] d x$

We know

$\boxed{\tt{ \: {tan}^{ - 1} \frac{x + y}{1 - xy} = {tan}^{ - 1}x + {tan}^{ - 1}y \: }}$

So, using this identity, we get

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: [{tan}^{ - 1}(1 - x) + {tan}^{ - 1}x] \: d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: {tan}^{ - 1}(1 - x) \: dx + \displaystyle \int_{0}^{1} \rm \:{tan}^{ - 1}x\: d x$

We know,

$\boxed{\tt{ \: \: \displaystyle \int_{0}^{a} \rm \:f(x) \: dx \: = \: \displaystyle \int_{0}^{a} \rm \:f(a - x) \: dx \: }}$

So, using this property, we get

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: {tan}^{ - 1}[1 - (1 - x)] \: dx + \displaystyle \int_{0}^{1} \rm \:{tan}^{ - 1}x\: d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: {tan}^{ - 1}[1 - 1 + x] \: dx + \displaystyle \int_{0}^{1} \rm \:{tan}^{ - 1}x\: d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: {tan}^{ - 1}x \: dx + \displaystyle \int_{0}^{1} \rm \:{tan}^{ - 1}x\: d x$

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: 2{tan}^{ - 1}x \: dx$

Now, Consider

$\rm \: \displaystyle \int \rm \: 2{tan}^{ - 1}x \: dx$

Using Integration by parts, we get

$\rm \: = \: {tan}^{ - 1}x\displaystyle \int\rm \: 2 \: dx \: - \: \displaystyle \int\rm \: \bigg[\dfrac{d}{dx}{tan}^{ - 1}x \: \displaystyle \int\rm \: 2 \: dx \bigg]dx$

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

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

We know,

$\boxed{\tt{ \displaystyle \int\rm \: \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: }}$

So, using this result, we get

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

So,

$\rm \: = \: \displaystyle \int_{0}^{1} \rm \: 2{tan}^{ - 1}x \: dx$

$\rm \: = \: \bigg |2x{tan}^{ - 1}x - log | {x}^{2} + 1 |\bigg |_{0}^{1}$

$\rm \: = \: 2{tan}^{ - 1}1 - log |2| - 0 + 0$

$\rm \: = \: 2 \times \dfrac{\pi}{4} - log |2|$

$\rm \: = \: \dfrac{\pi}{2} - log 2$

Hence,

$\boxed{\tt{ \rm \:\displaystyle \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right)\tt d x = \: \dfrac{\pi}{2} - log 2 \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: \: \displaystyle \int_{a}^{b} \rm \:f(x) \: dx \: = \: - \: \displaystyle \int_{b}^{a} \rm \:f(x) \: dx \: }}$

$\boxed{\tt{ \: \: \displaystyle \int_{a}^{b} \rm \:f(x) \: dx \: = \: \displaystyle \int_{a}^{b} \rm \:f(y) \: dy \: }}$

$\boxed{\tt{ \: \: \displaystyle \int_{a}^{b} \rm \:f(x) \: dx \: = \: \displaystyle \int_{a}^{b} \rm \:f(a + b - x) \: dx \: }}$

$\boxed{\tt{ \: \: \displaystyle \int_{ - a}^{a} \rm \:f(x) \: dx \: = 0 \: \: if \: f( - x) \: = \: - \: f(x)}}$

$\boxed{\tt{ \: \: \displaystyle \int_{ - a}^{a} \rm \:f(x) \: dx \: =2 \displaystyle \int_{0}^{a} \rm \:f(x)dx \: \: \: if \: f( - x) \: = f(x)}}$

$\boxed{\tt{ \: \: \displaystyle \int_{0}^{2a} \rm \:f(x) \: dx \: = 0 \: \: if \: f(2a - x) \: = \: - \: f(x)}}$

Integration by parts Formula

$\boxed{\tt{ \rm \: \displaystyle \int\rm \: uv \: dx = \: u\displaystyle \int\rm \: v \: dx \: - \: \displaystyle \int\rm \: \bigg[\dfrac{d}{dx}u \: \displaystyle \int\rm \: v \: dx \bigg]dx}}$

Answer 2

Question:-

$\sf \large{Evaluate:- \: \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x}.$

Given:-

$\sf \large{ \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x}.$

Solution:-

$\sf \large \int ^{1} _{0}(1 - x + x {}^{2} )dx$

$\sf \large = \int^{1} _{0}tan {}^{ - 1} \bigg \{ \frac{(1 - x) + x}{1 - x(1 - x)} \bigg \} dx$

$\sf \large = \int ^{1} _{0}tan {}^{ - 1} (1 - x)dx + \int ^{1} _{0}tan {}^{ - 1} \: x \: dx$

$\sf \large = - \bigg[(1 - x)tan {}^{ - 1}(1 - x) - \frac{1}{2}log \big \{1 + (1 - x) {}^{2} \big \} \bigg ]^{1} _{0} + \bigg \{x \: tan {}^{ - 1} x - \frac{1}{2} log(1 + x {}^{2} ) \bigg \} ^{1} _{0}$

$\sf \large = \frac{\pi}{2} - log2.$

Answer:-

$\sf \large {{\int^{1}_{0} cot {}^{ - 1} (1 - x + x {}^{2})dx \sf \large \: = \frac{\pi}{2} - log2 .}}$

Hope you have satisfied. ⚘