Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π






∫
 = 
2
2
1
– 0
2 16
32


π
=π




EXERCISE 7 10
Evaluate the integrals in Exercises 1 to 8 using substitution 1 1
2
0
1
x
dx
∫x +
2