Question

Evaluate : $\int _ { 0 } ^ { 1 } \cot ^ { - 1 }$(1 – x + x²) dx -

• A. $\frac { \pi } { 2 }$ – ln 2
• B. $\frac { \pi } { 2 }$ – ln 4
• C. $\frac { \pi } { 4 }$ – ln 2
• D. None

Answer 1

Answer: (A)

$\frac { \pi } { 2 }$ – ln 2

Answer 2

Let I = $\int _ { 0 } ^ { 1 } \cot ^ { - 1 }$ (1 – x + x²) dx

= $\int _ { 0 } ^ { 1 } \cot ^ { - 1 }$(1 – x (1 – x)) dx

(Q 0 £ x < 1)

= $\int _ { 0 } ^ { 1 } \tan ^ { - 1 }$dx

= $\int _ { 0 } ^ { 1 } \tan ^ { - 1 }$dx

= $\int _ { 0 } ^ { 1 } \left( \tan ^ { - 1 } x + \tan ^ { - 1 } ( 1 - x ) \right)$dx

=

(From property)

= 2dx

Integrating by parts taking unity as the second function, we have

I = 2

= 2

= 2 $\left[ \frac { \pi } { 4 } - \frac { 1 } { 2 } \ln 2 \right]$

Hence I =$\frac { \pi } { 2 }$ – ln 2.