Evaluate: $\int_{0}^{1} \cot^{-1}\left[1-x+x^{2}\right] d x$

CBSE Class 12 · 2008 · Definite IntegralsLong Answer

Evaluate:

\int_{0}^{1} cot^{- 1} [1 - x + x^{2}] d x

Answer: $\frac{π}{2} - lo g 2$

Write

cot^{- 1} (1 - x + x^{2}) = tan^{- 1} \frac{1}{1 - x + x ^{2}} = tan^{- 1} \frac{x + ( 1 - x )}{1 - x ( 1 - x )} = tan^{- 1} x + tan^{- 1} (1 - x)

I = \int_{0}^{1} tan^{- 1} x d x + \int_{0}^{1} tan^{- 1} (1 - x) d x

In the second integral

x \to 1 - x

turns it into

\int_{0}^{1} tan^{- 1} x d x

, so

I = 2 \int_{0}^{1} tan^{- 1} x d x

.
Integrate by parts:

\int_{0}^{1} tan^{- 1} x d x = [x tan^{- 1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x ^{2}} d x = \frac{π}{4} - \frac{1}{2} lo g 2

I = 2 (\frac{π}{4} - \frac{1}{2} lo g 2) = \frac{π}{2} - lo g 2

CBSE Class 12 Board · 2008 · Subjective