PRELIMINARY CALCULUS
Sincedt
dx=1
2sec^2x
2=1
2(
1+tan^2x
2)
=1+t^2
2,the required relationship is
dx=2
1+t^2dt. (2.34)Evaluate the integralI=∫
2
1+3cosxdx.Rewriting cosxin terms oftand using (2.34) yields
I=
∫
2
1+3
[
(1−t^2 )(1 +t^2 )−^1]
(
2
1+t^2)
dt=
∫
2(1 +t^2 )
1+t^2 +3(1−t^2 )(
2
1+t^2)
dt=
∫
2
2 −t^2dt=∫
2
(
√
2 −t)(√
2+t)dt=
∫
1
√
2
(
1
√
2 −t+
1
√
2+t)
dt=−
1
√
2
ln(√
2 −t)+1
√
2
ln(√
2+t)+c=
1
√
2
ln[√
2+tan(x/2)
√
2 −tan (x/2)]
+c.Integrals of a similar form to (2.33), but involving sin 2x,cos2x, tan 2x,sin^2 x,cos^2 xor tan^2 xinstead of cosxand sinx, should be evaluated by using the
substitutiont=tanx.Inthiscase
sinx=t
√
1+t^2, cosx=1
√
1+t^2and dx=dt
1+t^2. (2.35)
A final example of the evaluation of integrals using substitution is the methodof completing the square (cf. subsection 1.7.3).
