Evaluate
(i) 
(ii) 
Text Solution
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Sol. We have,
= 
= 
= 
= 
= 
+ 
= I 1 + I 2 , where ... (i)
I 1 = 
and I 2 = 
Now, I 1 = 
= 
= x tan –1 x – 
dx
= x tan –1 x – 
= x tan –1 x – 
log (1+ x 2 ) ... (ii) and, I 2 = 
= – 
= – 
[Using ii)]
Substituting the values of I 1 and I 2 in (i), we get
= x tan –1 x– 
log (1+ x 2 ) – (1–x)
tan –1 (1 –x)+ 
log {1+ (1–x) 2 +C
(ii) Let I = 
. sec 2 θ d θ . Then,
I = 
sec 2 θ d θ
= 
sec 2 θ d θ
= 
, where t = tan θ
= 
= 
{tan –1 t + tan –1 (1 –t)} dt
= t tan –1 t – 
log (1+ t 2 ) – (1 –t) tan –1 (1 –t) + 
log {1 + (1 –t) 2 + C} [Using (i)]
= θ tan θ + log cos θ – (1 –tan θ ) tan –1 (1 –tan θ ) + 
log (2 –2 tan θ + tan 2 θ ) + C
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