Use the substitute x= tan y to show that

Question

∫¹       1     dx =¶  +  1

0  (1+x²)²         8      4

Answer 1

x = tany => x = 0, y = 0; x = 1, y = pi/4

dx = sec²y dy

∫{0,1}1/(1+x²)² dx

= ∫{0, pi/4} cos²y dy

= ∫{0, pi/4} (1/2)(1+cos2y) dy

= (1/2)y + (1/4)sin(2y) from 0 to pi/4

= pi/8 + 1/4, since sin(2*pi/4) = 1.

Answer 2

Use identity 1/(1+tan(x)²)=cos (x)² Your integral becomes, upon substitution x=tan (y),

∫₀¹ dx/(1+x²)²=∫₀^arctan(1) d(tan (y))*cos (y)⁴= ∫₀^arctan(1) dy/(cos(y))²*cos(y)⁴=∫₀^arctan(1) dy * cos(y)²=

=½∫₀^arctan(1) dy (1+cos(2y))=[y/2+¼sin(2y)] |₀^arctan(1)=

arctan(1)/2+¼sin(2*arctan(1))=arctan(1)/2+¼;

Here I used the fact that sin(2 arctan(y))=2y/(1+y²); arctan(1)=pi/4, hence

 

Answer: pi/8+1/4