10th lecture,,, Integration 14 09 2020 PDF

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Anti-differentiation Reduction formula: The reduction formula for ∫ sinn 𝑥 𝑑𝑥 Let 𝐼𝑛 = ∫ sin𝑛 𝑥 𝑑𝑥 = ∫ sin𝑛−1 𝑥 sin 𝑥 𝑑𝑥 = sin𝑛−1 𝑥 ∫ sin 𝑥 𝑑𝑥 − ∫{

𝑑 (sin𝑛−1 𝑥) ∫ sin 𝑥 𝑑𝑥 }𝑑𝑥 𝑑𝑥

= − sinn−1 cos 𝑥 − ∫(𝑛 − 1) sin𝑛−2 𝑥 cos 𝑥 (− cos 𝑥)𝑑𝑥 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1) ∫ sin𝑛−2 𝑥 cos2 𝑥 𝑑𝑥 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1) ∫ sin𝑛−2 𝑥 (1 − sin2 𝑥)𝑑𝑥 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1) ∫ sinn−2 𝑥 𝑑𝑥 − (𝑛 − 1) ∫ sin𝑛 𝑥 𝑑𝑥 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1)𝐼𝑛−2 − (𝑛 − 1)𝐼𝑛

 𝐼𝑛 + (𝑛 − 1)𝐼𝑛 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1)𝐼𝑛−2  𝑛 𝐼𝑛 = − sin𝑛−1 𝑥 cos 𝑥 + (𝑛 − 1)𝐼𝑛−2 1

 𝐼𝑛 = − 𝑛 sin𝑛−1 𝑥 cos 𝑥 +

Example: Evaluate ∫ sin6 𝑥 𝑑𝑥

(𝑛−1) 𝑛

𝐼𝑛−2 , 𝑛 ≥ 2

∴ 𝐼6 = ∫ sin6 𝑥 𝑑𝑥 1 6−1 𝐼 = − sin6−1 𝑥 cos 𝑥 + 6 4 6 1 3 5 1 = − sin5 𝑥 cos 𝑥 + [− sin3 𝑥 cos 𝑥 + 𝐼2 ] 6 4 6 4

1 1 5 1 5 = − sin5 𝑥 cos 𝑥 − sin3 𝑥 cos 𝑥 + [− sin0 𝑥 cos 𝑥 + 𝐼0 ] 8 2 24 6 2 1 5 5 5 = − sin5 𝑥 cos 𝑥 − cos 𝑥 + ∫ sin0 𝑥 𝑑𝑥 sin3 𝑥 cos 𝑥 − 16 24 6 16 1 5 5 5 == − sin5 𝑥 cos 𝑥 − cos 𝑥 + ∫ 𝑑𝑥 sin3 𝑥 cos 𝑥 − 16 24 6 16 1 5 5 5 = − sin5 𝑥 cos 𝑥 − cos 𝑥 + 𝑥 + 𝑐 sin3 𝑥 cos 𝑥 − 16 24 6 16

H. W.: (1) Obtain the reduction formula for ∫ cos𝑛 𝑥 𝑑𝑥 and hence evaluate ∫ cos5 𝑥 𝑑𝑥. (2) Obtain the reduction formula for ∫ tan𝑛 𝑥 𝑑𝑥 and hence evaluate ∫ tan5 𝑥 𝑑𝑥.

(3) Obtain the reduction formula for ∫ sec 𝑛 𝑥 𝑑𝑥 and hence evaluate ∫ sec 5 𝑥 𝑑𝑥. Ans. (3): let 𝐼𝑛 = ∫ sec𝑛 𝑥 𝑑𝑥

= ∫ sec 𝑛−2 𝑥 sec 2 𝑥 𝑑𝑥 𝑑

= sec 𝑛−2 𝑥 ∫ sec 2 𝑥 𝑑𝑥 − ∫{ 𝑑𝑥 (sec 𝑛−2 𝑥) ∫ sec2 𝑥 𝑑𝑥 }𝑑𝑥 = sec 𝑛−2 𝑥 tan 𝑥 − ∫(𝑛 − 2) sec 𝑛−3 𝑥 sec 𝑥 tan 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑛−2 𝑥 tan 𝑥 − (𝑛 − 2) ∫ sec n−2 𝑥 tan2 𝑥 𝑑𝑥 = sec 𝑛−2 𝑥 tan 𝑥 − (𝑛 − 2) ∫ sec 𝑛−2 𝑥 (sec2 𝑥 − 1)𝑑𝑥 = sec 𝑛−2 𝑥 tan 𝑥 − (𝑛 − 2) ∫ sec 𝑛 𝑥 𝑑𝑥 + (𝑛 − 2) ∫ sec n−2 𝑥 𝑑𝑥 = sec 𝑛−2 𝑥 tan 𝑥 − (𝑛 − 2)𝐼𝑛 + (𝑛 − 2)𝐼𝑛−2

 𝐼𝑛 (1 + 𝑛 − 2) = sec 𝑛−2 𝑥 tan 𝑥 + (𝑛 − 2)𝐼𝑛−2 𝑛−2 1  𝐼𝑛 = 𝑛−1 sec𝑛−2 𝑥 tan 𝑥 + 𝑛−1 𝐼𝑛−2 ; 𝑛 ≥ 2 ∴ 𝐼5 = ∫ sec 5 𝑥 𝑑𝑥 =

1

1 5−2 𝐼 sec 5−2 𝑥 tan 𝑥 + 5−1 3 5−1

1 1 3 1 = sec 3 𝑥 tan 𝑥 + [ sec 𝑥 tan 𝑥 + 𝐼1 ] 4 2 4 2 1 3 3 = sec 3 𝑥 tan 𝑥 + sec 𝑥 tan 𝑥 + ∫ sec 𝑥 𝑑𝑥 8 4 8 3

3

= sec 3 𝑥 tan 𝑥 + sec 𝑥 tan 𝑥 + ln | sec 𝑥 + tan 𝑥 | + 𝑐 8 4 8 Exercise 7.3 (1-50)

Integrating product of 𝒔𝒊𝒏𝒆𝒔 and 𝒄𝒐𝒔𝒊𝒏𝒆𝒔: ∫ sinm 𝑥 cos𝑛 𝑥 𝑑𝑥 𝑛 odd

𝑚 odd

Split off a factor of cos 𝑥, use the identity cos 2 𝑥 = 1 − sin2 𝑥 then substitute 𝑢 = sin 𝑥 Split off a factor of sin 𝑥, use the identity sin2 𝑥 = 1 − cos 2 𝑥 then substitute 𝑢 = cos 𝑥

𝑚, 𝑛 Use sin2 𝑥 = 1 (1 − cos 2𝑥) 2 both 2 𝑥 = 1 (1 + cos 2𝑥), and cos even 2 substitute 𝑢 = 2𝑥 , then use reduction formula.

∫ sin4 𝑥 cos 3 𝑥 𝑑𝑥 4

= ∫ sin 𝑥 cos2 𝑥 cos 𝑥 𝑑𝑥

𝑢 = sin 𝑥 𝑑𝑢 = cos 𝑥 𝑑𝑥

= ∫ sin4 𝑥 (1 − sin2 𝑥) cos 𝑥 𝑑𝑥

= ∫ 𝑢4 (1 − 𝑢2 )𝑑𝑢…

∫ sin3 𝑥 cos 4 𝑥 𝑑𝑥 2

= ∫ cos4 𝑥 sin 𝑥 sin 𝑥 𝑑𝑥

𝑢 = cos 𝑥 𝑑𝑢 = − sin 𝑥

= ∫ cos 4 𝑥 (1 − cos 2 𝑥) sin 𝑥 𝑑𝑥

= − ∫ 𝑢4 (1 − 𝑢2 )𝑑𝑢…

∫ sin6 𝑥 cos 6 𝑥 𝑑𝑥

3 1 1 = ∫ { (1 − cos 2𝑥)} { (1 2 2 3

+ cos 2𝑥 )} 𝑑𝑥

𝑢 = 2𝑥 𝑑𝑢 = 2 𝑑𝑥 1 𝑑𝑥 = 𝑑𝑢 2

1 3 ∫(1 − cos2 2𝑥) 𝑑𝑥 64 3 1 2 ∫ (sin 2𝑥) 𝑑𝑥 = 64 1 6 ∫ sin 2𝑥 𝑑𝑥 = 64 1 6 = 128 ∫ sin 𝑢 𝑑𝑢… =

 Integrals of the form ∫ 𝑠𝑖𝑛 𝑚𝑥 𝑐𝑜𝑠 𝑛𝑥 𝑑𝑥, ∫ 𝑠𝑖𝑛 𝑚𝑥 𝑠𝑖𝑛 𝑛𝑥 𝑑𝑥, ∫ 𝑐𝑜𝑠 𝑚𝑥 𝑐𝑜𝑠 𝑛𝑥 𝑑𝑥 can be found by using the trigonometric identities 𝑠𝑖𝑛 𝛼 𝑐𝑜𝑠 𝛽 = 𝑠𝑖𝑛 𝛼 𝑠𝑖𝑛 𝛽 =

1

2 1

[𝑠𝑖𝑛 (𝛼 − 𝛽) + 𝑠𝑖𝑛 (𝛼 + 𝛽)]

[𝑐𝑜𝑠 (𝛼 − 𝛽) − 𝑐𝑜𝑠 (𝛼 + 𝛽)] 1 𝑐𝑜𝑠 𝛼 𝑐𝑜𝑠 𝛽 = [𝑐𝑜𝑠 (𝛼 − 𝛽) + 𝑐𝑜𝑠 (𝛼 + 𝛽)] 2 to express the integrand as a sum or difference of sines and cosine. 2

Example: Evaluate ∫ sin 7𝑥 cos 3𝑥 𝑑𝑥 1

Ans. ∫ sin 7𝑥 cos 3𝑥 𝑑𝑥 = ∫ [sin(7𝑥 + 3𝑥) + sin(7𝑥 − 3𝑥)] 2

1 2

=−

∫(sin 10𝑥 + sin 4𝑥)𝑑𝑥 1

1 cos 10𝑥 − 8 cos 4𝑥 + 𝑐

20 Integrating product of 𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒔 and 𝒔𝒆𝒄𝒂𝒏𝒕𝒔: ∫ tanm 𝑥 sec 𝑛 𝑥 𝑑𝑥 𝑛 even

𝑚 odd

Split off a factor of sec2 𝑥, use identity sec2 𝑥 = 1 + tan2 𝑥, then substitute 𝑢 = tan 𝑥

∫ tan3 𝑥 sec 4 𝑥 𝑑𝑥 = ∫ tan3 𝑥 sec2 𝑥 sec2 𝑥 𝑑𝑥

𝑢 = tan 𝑥 𝑑𝑢 = sec 2 𝑥 𝑑𝑥

= ∫ tan3 𝑥(1 + tan2 𝑥) sec 2 𝑥 𝑑𝑥

= ∫ 𝑢 3 (1 + 𝑢 2 )𝑑𝑢 …

Split off a factor of ∫ tan5 𝑥 sec 3 𝑥 𝑑𝑥 𝑠𝑒𝑐 𝑥 tan 𝑥, use identity tan2 𝑥 = = ∫ tan4 𝑥 sec2 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥 2 sec 𝑥 − 1, then substitute 𝑢 = sec 𝑥 = ∫(sec2 𝑥

𝑢 = sec 𝑥

𝑑𝑢 = sec 𝑥 tan 𝑥 𝑑𝑥

− 1)2 sec2 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥 = ∫(𝑢 2 − 1)2 𝑢 2 𝑑𝑢…

𝑛 odd Express the integrand ∫ tan4 𝑥 sec 𝑥 𝑑𝑥 in terms of sec 𝑥 and 𝑚 even use the reduction = ∫(sec2 𝑥 − 1)2 sec 𝑥 𝑑𝑥 formula of sec 𝑥

= ∫(sec 5 𝑥 − 2 sec 3 𝑥 + sec 𝑥) 𝑑𝑥 = ∫ sec5 𝑥 𝑑𝑥 − 2 ∫ sec 3 𝑥 𝑑𝑥 + ∫ sec 𝑥 𝑑𝑥…

 If ∫ cot m 𝑥 cosec 𝑛 𝑥 𝑑𝑥, then we can use the same method as ∫ tanm 𝑥 sec 𝑛 𝑥 𝑑𝑥...