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Question: Evaluate \[\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx} \] A.\[\pi - \dfrac{{\log }}{2}\] ...

Evaluate ∫01cot⁡−1(1−x+x2)dx\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx}
A.π−log⁡2\pi - \dfrac{{\log }}{2}
B. 2π−log⁡22\pi - \dfrac{{\log }}{2}
C. π+log⁡2\pi + \dfrac{{\log }}{2}
D. None of these

Explanation

Solution

We use the property that tangent and cotangent functions are reciprocal of each other. Convert the integral in terms of inverse tangent function and break the numerator in such a way that we can apply the property of addition of two tangent inverse functions. Use integration by parts to find the value of the obtained integral.

  • ∵cot⁡θ=1tan⁡θ⇒cot⁡−1θ=tan⁡−11θ\because \cot \theta = \dfrac{1}{{\tan \theta }} \Rightarrow {\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }
  • tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)
    *In integration by parts we use the sequence of ILATE (Inverse trigonometric, Logarithms, algebraic, trigonometric, and exponential) for choosing the value of u,vu,v and then substitute in the integration formula of by parts integration.

Complete step by step solution:
We have to evaluate the integral ∫01cot⁡−1(1−x+x2)dx\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx}
Convert the integral in terms of tangent inverse
⇒cot⁡−1(1−x+x2)dx=∫01tan⁡−11(1−x+x2)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{1}{{(1 - x + {x^2})}}} dx................… (1)
Now we break the numerator in such a way so we can apply property of trigonometry
We can write 1=x+1−x1 = x + 1 - x
Substitute the value of 1=x+1−x1 = x + 1 - x in numerator of equation (1)
⇒cot⁡−1(1−x+x2)dx=∫01tan⁡−1x+1−x(1−x+x2)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{(1 - x + {x^2})}}} dx
Write the denominator in easy terms
⇒cot⁡−1(1−x+x2)dx=∫01tan⁡−1x+1−x1−x(1−x)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{1 - x(1 - x)}}} dx...............… (2)
If we take two angles, (x)(x) and (1−x)(1 - x) then we can write tan⁡−1(x)+tan⁡−1(1−x)=tan⁡−1(x+(1−x)1−x(1−x)){\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(1 - x) = {\tan ^{ - 1}}\left( {\dfrac{{x + (1 - x)}}{{1 - x(1 - x)}}} \right)
So, RHS of the equation (2) transforms into easier form using the identity tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)
⇒cot⁡−1(1−x+x2)dx=∫01(tan⁡−1(x)+tan⁡−1(1−x))dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {\left( {{{\tan }^{ - 1}}(x) + {{\tan }^{ - 1}}(1 - x)} \right)dx}
Separate the two inverse functions in RHS
⇒cot⁡−1(1−x+x2)dx=∫01tan⁡−1(x)dx+∫01tan⁡−1(1−x)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx}.................… (3)
Since we know that ∫0af(x)dx=∫0af(a−x)dx\int\limits_0^a {f(x)} dx = \int\limits_0^a {f(a - x)} dx
So, ∫01tan⁡−1(1−x)dx=∫01tan⁡−11−(1−x)dx\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}1 - (1 - x)dx}
Which on solving gives ∫01tan⁡−1(1−x)dx=∫01tan⁡−1(1−1+x)dx\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - 1 + x} \right)dx}
i.e. ∫01tan⁡−1(1−x)dx=∫01tan⁡−1xdx\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx}
Substitute the value of ∫01tan⁡−1(1−x)dx=∫01tan⁡−1xdx\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx} in equation (3)
⇒cot⁡−1(1−x+x2)dx=∫01tan⁡−1(x)dx+∫01tan⁡−1(x)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx}
Add like terms in RHS
⇒cot⁡−1(1−x+x2)dx=2∫01tan⁡−1(x)dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}(x)dx}
Now we can write RHS as
⇒cot⁡−1(1−x+x2)dx=2∫01tan⁡−1x×1dx\Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx}
Use by parts to integrate RHS
Looking at the integral 2∫01tan⁡−1x×1dx2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx}
Using the ILATE sequence we can tell that the term comes first in the sequence is uu and which comes later is vv i.e. in this case we have the terms inverse trigonometric and algebraic , first comes the inverse trigonometric term then comes the algebraic term so uu is tan⁡−1x{\tan ^{ - 1}}x and vv is 1
u=tan⁡−1x,v=1u = {\tan ^{ - 1}}x,v = 1
Now we calculate the differentiation of the term uu
u′=dudxu' = \dfrac{{du}}{{dx}}
⇒u′=d(tan⁡−1x)dx\Rightarrow u' = \dfrac{{d({{\tan }^{ - 1}}x)}}{{dx}}
⇒u′=11+x2\Rightarrow u' = \dfrac{1}{{1 + {x^2}}}
Now substitute the values in equation ∫uvdt=u∫vdt−∫u′(∫vdt)dt\int {uvdt = u\int {vdt - \int {u'\left( {\int {vdt} } \right)dt} } }
⇒2∫01tan⁡−1x.1dx=2tan⁡−1x∫011dx−2∫01(11+x2)(∫1dx)dx\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x\int\limits_0^1 {1dx} - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})\left( {\int {1dx} } \right)dx}
Now we know ∫1dx=x\int {1dx = x}
Therefore substitute the value of ∫1dx=x\int {1dx = x} in RHS
⇒2∫01tan⁡−1x.1dx=2tan⁡−1x(x)−2∫01(11+x2)(x)dx\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x(x) - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})(x)dx}
Now we can solve and bring out constant terms from the integral sign.
⇒2∫01tan⁡−1x.1dx=2xtan⁡−1x−∫01(2x1+x2)dx\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2x{\tan ^{ - 1}}x - \int\limits_0^1 {(\dfrac{{2x}}{{1 + {x^2}}})dx}
Since the numerator of the fraction in RHS is derivative of the denominator and we know ∫dtt=log⁡t\int {\dfrac{{dt}}{t} = \log t}
⇒2∫01tan⁡−1x.1dx=2[xtan⁡−1x]01−[log⁡(1+x2)]01\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {x{{\tan }^{ - 1}}x} \right]_0^1 - \left[ {\log \left( {1 + {x^2}} \right)} \right]_0^1
Calculate the values by substituting the upper and lower limits
⇒2∫01tan⁡−1x.1dx=2[1tan⁡−11−0]−[log⁡(1+1)−log⁡(1+0)]\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {1{{\tan }^{ - 1}}1 - 0} \right] - \left[ {\log \left( {1 + 1} \right) - \log \left( {1 + 0} \right)} \right]
⇒2∫01tan⁡−1x.1dx=2[tan⁡−11]−[log⁡2−log⁡1]\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right]
Substitute tan⁡−11=π4{\tan ^{ - 1}}1 = \dfrac{\pi }{4} and use the value log⁡1=0\log 1 = 0
⇒2∫01tan⁡−1x.1dx=2×π4−[log⁡2−0]\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2 \times \dfrac{\pi }{4} - \left[ {\log 2 - 0} \right]
⇒2∫01tan⁡−1x.1dx=π2−log⁡2\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2
So value of 2∫01tan⁡−1x.1dx2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} is π2−log⁡2\dfrac{\pi }{2} - \log 2.

∴\therefore Option D is correct.

Note: Alternative method:
We can solve the solution after the step 2∫01tan⁡−1x.1dx=2[tan⁡−11]−[log⁡2−log⁡1]2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right] in another way
For step in the end we can use rule of log i.e. log⁡m−log⁡n=log⁡mn\log m - \log n = \log \dfrac{m}{n}
⇒2∫01tan⁡−1x.1dx=2[tan⁡−11]−[log⁡21]\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log \dfrac{2}{1}} \right]
⇒2∫01tan⁡−1x.1dx=π2−log⁡2\Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2
So value of 2∫01tan⁡−1x.1dx2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} isπ2−log⁡2\dfrac{\pi }{2} - \log 2.
∴\therefore Option D is correct.