Integration using trigonometric and hyperbolic substitutions 399
Table 40.1Integrals using trigonometric and hyperbolic substitutions
f(x)∫
f(x)dx Method See problem- cos^2 x
1
2(
x+sin2x
2)
+c Use cos2x=2cos^2 x− 1 1- sin^2 x
1
2(
x−sin2x
2)
+c Use cos2x= 1 −2sin^2 x 2- tan^2 x tanx−x+c Use 1+tan^2 x=sec^2 x 3
- cot^2 x −cotx−x+c Use cot^2 x+ 1 =cosec^2 x 4
- cosmxsinnx (a) If eithermornis odd (but not both), use
cos^2 x+sin^2 x= 1 5, 6
(b)Ifbothmandnare even, use either
cos 2x=2cos^2 x−1orcos2x= 1 −2sin^2 x 7, 8 - sinAcosB Use^12 [sin(A+B)+sin(A−B)] 9
- cosAsinB Use^12 [sin(A+B)−sin(A−B)] 10
- cosAcosB Use^12 [cos(A+B)+cos(A−B)] 11
- sinAsinB Use−^12 [cos(A+B)−cos(A−B)] 12
10.1
√
(a^2 −x^2 )sin−^1x
a+c Usex=asinθsubstitution 13, 1411.√
(a^2 −x^2 )a^2
2sin−^1x
a+x
2√
(a^2 −x^2 )+c Usex=asinθsubstitution 15, 1612.1
a^2 +x^21
atan−^1x
a+c Usex=atanθsubstitution 17–1913.1
√
(x^2 +a^2 )sinh−^1x
a+c Usex=asinhθsubstitution 20–22or ln{
x+√
(x^2 +a^2 )
a}
+c14.√
(x^2 +a^2 )a^2
2sinh−^1x
a+x
2√
(x^2 +a^2 )+c Usex=asinhθsubstitution 2315.1
√
(x^2 −a^2 )cosh−^1x
a+c Usex=acoshθsubstitution 24, 25or ln{
x+√
(x^2 −a^2 )
a}
+c16.√
(x^2 −a^2 )x
2√
(x^2 −a^2 )−a^2
2cosh−^1x
a+c Usex=acoshθsubstitution 26, 27