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Question: Find the Integration of - \(\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 + x + {x^2}} \right)dx} \)...

Find the Integration of - ∫01cot⁡−1(1+x+x2)dx\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 + x + {x^2}} \right)dx}

Explanation

Solution

Hint: To solve this question first we use ( cot⁡−1x=tan⁡−11x{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}) and then using the formula of inverse trigonometric function (tan⁡−1(x−y1+xy)=tan⁡−1x−tan⁡−1y{\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y) using this property we can easily solve the question further.

Complete step-by-step answer:
We have given in the question:
∫01cot⁡−1(1+x+x2)dx\int\limits_0^1 {{{\cot }^{ - 1}}\left( {1 + x + {x^2}} \right)dx}
Now using cot⁡−1x=tan⁡−11x{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x} this property we can write it as
∫01tan⁡−1(11+x+x2)dx\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + x + {x^2}}}} \right)dx}
We can write it as also
∫01tan⁡−1(11+x(1+x))dx\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{1}{{1 + x\left( {1 + x} \right)}}} \right)dx}
To apply property we will write it as :
∫01tan⁡−1((1+x)−x1+x(1+x))dx\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{{\left( {1 + x} \right) - x}}{{1 + x\left( {1 + x} \right)}}} \right)dx}
Now we can apply the property tan⁡−1(x−y1+xy)=tan⁡−1x−tan⁡−1y{\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}x - {\tan ^{ - 1}}y using this property we can break it into two parts.
∫01[(tan⁡−1(1+x))−tan⁡−1x]dx\int\limits_0^1 {\left[ {\left( {{{\tan }^{ - 1}}\left( {1 + x} \right)} \right) - {{\tan }^{ - 1}}x} \right]dx}
Now for further integration we will use integration by parts:
We will write it as
∫011.tan⁡−1(1+x)dx−∫011.tan⁡−1(x)dx\int\limits_0^1 {1.{{\tan }^{ - 1}}\left( {1 + x} \right)dx} - \int\limits_0^1 {1.{{\tan }^{ - 1}}\left( x \right)dx}
[xtan⁡−1(x+1)]10−∫01x1+x2dx−[xtan⁡−1x−∫011x2.xdx]10{\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - \int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} - {\left[ {x{{\tan }^{ - 1}}x - \int\limits_0^1 {\dfrac{1}{{{x^2}}}.xdx} } \right]^1}_0
[xtan⁡−1(x+1)]10−∫01x1+x2dx−[xtan⁡−1x−∫01dxx]10{\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - \int\limits_0^1 {\dfrac{x}{{1 + {x^2}}}dx} - {\left[ {x{{\tan }^{ - 1}}x - \int\limits_0^1 {\dfrac{{dx}}{x}} } \right]^1}_0
Now on integrating we get,
[xtan⁡−1(x+1)]10−[12log⁡(1+x2)]10−[xtan⁡−1x]10+[log⁡x]10{\left[ {x{{\tan }^{ - 1}}\left( {x + 1} \right)} \right]^1}_0 - {\left[ {\dfrac{1}{2}\log \left( {1 + {x^2}} \right)} \right]^1}_0 - {\left[ {x{{\tan }^{ - 1}}x} \right]^1}_0 + {\left[ {\log x} \right]^1}_0
On putting the limit we get,
tan⁡−12−12log⁡2−π4{\tan ^{ - 1}}2 - \dfrac{1}{2}\log 2 - \dfrac{\pi }{4}

Note:- Whenever we get this type of question the key concept of solving is we have to have knowledge of integration and remember formula of integration by parts that is ∫uvdx=u∫vdx−∫(dudx∫vdx)dx\int {uvdx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)dx} } }
Because this is the most important method for integration.