Integration of inverse trig. function

Calculus Level 4

Given : ∫ 0 1 cot ⁡ − 1 ( 1 − x + x 2 ) d x = a π − ln ⁡ λ \displaystyle \int_{0}^1 \cot^{-1} (1-x+x^2) \text{ }dx =a \pi-\ln \lambda

Find the value of a + λ a + \lambda .


The answer is 2.5.

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Sandeep Rathod
Nov 24, 2014

∫ 0 1 t a n − 1 1 1 − x + x 2 \int_{0}^{1} tan^{-1} \frac{1}{1 - x + x^{2}}

= ∫ 0 1 t a n − 1 x − ( x − 1 ) 1 + x ( x − 1 ) = \int_{0}^{1} tan^{-1} \frac{x - (x - 1)}{1 + x(x - 1)}

= ∫ 0 1 t a n − 1 x − ∫ 0 1 t a n − 1 ( x − 1 ) = \int_{0}^{1}tan^{-1}x - \int_{0}^{1}tan^{-1}(x - 1)

= ∫ 0 1 t a n − 1 x − ∫ 0 1 t a n − 1 ( ( 1 − x ) − 1 ) = \int_{0}^{1} tan^{-1}x - \int_{0}^{1} tan^{-1}(( 1-x) -1)

= 2 ∫ 0 1 t a n − 1 x d x = 2\int_{0}^{1}tan^{-1}xdx

Applying by parts and keeping the limits , we get

= 2 [ x t a n − 1 − 1 2 l o g ( 1 + x 2 ) ] 0 1 = 2[xtan^{-1} - \frac{1}{2}log(1 + x^{2})]_{0}^{1}

= π 2 − l o g 2 = \frac{\pi}{2} - log2

12 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution.!

Sandeep Bhardwaj - 6 years, 6 months ago

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