4 - This module is about integration leading to Inverse Trigonometric Functions PDF

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Summary

This module is about integration leading to Inverse Trigonometric Functions such as arcsin x, arctan x, and arcsec x.

NOTE:
THIS IS A LECTURE NOTE OF AMA BLENDED LEARNING AND AMA ONLINE EDUCATION ...

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Integral Calculus Integration Leading to Inverse Trigonometric Functions

1

Integration Leading to Inverse Trigonometric Functions This module is about integration leading to Inverse Trigonometric Functions such as arcsin x, arctan x, and arcsec x. Course Module Objectives: At the end of this module, the learner should be able to: 1. Transform the integrand into form identical to the formula; 2. Apply the appropriate formula; and 3. Evaluate the integrand leading to inverse trigonometric function.

Basic Integration Formulas: Integration Yielding Inverse Trigonometric Functions 𝑢 𝑑𝑢 1. ∫ 2 2 = 𝑎𝑟𝑐𝑠𝑖𝑛 + c √𝑎 − 𝑢 𝑑𝑢

∫ 𝑢 2 + 𝑎2 =

2.

𝑑𝑢

𝑎

1

𝑎𝑟𝑐𝑡𝑎𝑛 𝑎 1

𝑢

𝑎

+𝑐 𝑢

∫ 𝑢√𝑢2 − 𝑎2 = 𝑎 𝑎𝑟𝑐𝑠𝑒𝑐 𝑎 + 𝑐 In dealing with these integrands, always remember to transform the given into u du form. Likewise, take note that du is the derivative of u. 3.

Illustrative Examples: 5𝑑𝑥

∫ 4+ 𝑥2

1.

Solution: Always check the denominator against the numerator. If u = 4 + x2; du = 2xdx which is not the case in this problem. If u = x; du = dx. Then think of the appropriate formula. It does not involve radical. Let a = 2; u = x; du = dx 𝑑𝑢 1 𝑢 ∫ 2 = arctan + 𝑐 2 𝑎 𝑎 + 𝑢 𝑎 𝟏 𝒙 Answer: = 𝟓 𝐚𝐫𝐜𝐭𝐚𝐧 + 𝒄 𝟐 𝟐 𝑑𝑥

∫ 4+ 9𝑥2

2.

Solution: Again, this does not involve radical. 𝑑𝑢

=∫ 𝑎 2 + 𝑢2 =

1

𝑢

arctan + 𝑐 𝑎 𝑎

Integral Calculus Integration Leading to Inverse Trigonometric Functions

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If u2 = x2; u = x; and du = dx. Transforming the integrand results to 1 𝑑𝑥 𝑑𝑥 ∫ 4 2 = 9 ∫ 22 9( 9+ 𝑥 )

( 3) + 𝑥 2

Which is similar to the formula 1 𝑑𝑢 1 1 𝑢 = ∫ 2 = ∗ arctan + 𝑐 2 𝑎 9 𝑎 + 𝑢 9 𝑎 1 1 𝑥 = ∗ 𝑎𝑟𝑐𝑡𝑎𝑛 2 + 𝐶 9 2⁄ 3 3 𝟑𝒙 𝟏 −𝟏 +𝑪 𝐭𝐚𝐧 = 𝟐 𝟔 3.

4𝑑𝑥

∫ √25− 𝑥2

Solution: This integrand involves radical. Check the terms a2 and u2. Compare with the formulas leading to arcsin u and arcsec u. 𝐿𝑒𝑡:

= ∫

𝑥 = 5𝑢, 4𝑑𝑥

𝑑𝑥 = 5𝑑𝑢 = 4∫

√25 − 25𝑢2 𝒙 = 𝟒 𝒂𝒓𝒄𝒔𝒊𝒏 + 𝑪 𝟓 4.

∫

𝑑𝑢

√𝑎2 − 𝑢 2

= 4 𝑎𝑟𝑐𝑠𝑖𝑛

𝑢 +𝐶 𝑎

𝑑𝑥

√25− 16𝑥 2

Solution: Check the denominator, and compare the formulas involving radicals. = ∫

𝑑𝑥

=

1 ∫ 4

𝑑𝑥

2 √16 (25 − 𝑥 2 ) √ (5 ) − 𝑥 2 16 4 1 𝑢 𝑑𝑢 1 ∫ = 𝑎𝑟𝑐𝑠𝑖𝑛 + 𝐶 𝑎 4 4 √𝑎2 − 𝑢 2 𝟏 𝟒𝒙 +𝑪 = 𝐚𝐫𝐜𝐬𝐢𝐧 𝟓 𝟒

5.

𝑑𝑥

∫ (𝑥−3)√𝑥2

=

1 𝑎𝑟𝑐𝑠𝑖𝑛 4

𝑥 +𝐶 5 4

−6𝑥+8

Solution: This problem involves variable outside the radical. Check the terms inside the radical and check if it can be transformed to the terms outside the radical.

Integral Calculus Integration Leading to Inverse Trigonometric Functions

𝐿𝑒𝑡: 𝑢 = 𝑥 − 3, 𝑑𝑢 = 𝑑𝑥, and a = 1 𝑥 2 − 6𝑥 + 8 = (𝑥 2 − 6𝑥 + 9) − 1 = (𝑥 − 3)2 − 1 = 𝑢 2 − 1 𝑑𝑢 𝑢 ∫ = sec−1 + 𝐶 𝑎 𝑢√𝑢2 − 𝑎2 −𝟏 (𝒙 = 𝐬𝐞𝐜 − 𝟑) + 𝑪 𝑑𝑥

∫ √3− 2𝑥2

6.

Solution: 𝑑𝑥

= ∫

=

1

∫

𝑑𝑥

√2 √3 2 √ 2 (3 − 𝑥 2 ) 2 2− 𝑥 1 𝑑𝑢 1 𝑥 = ∫ = sin−1 +𝐶 2 2 √2 √𝑎 − 𝑢 √2 3 √ 2 𝒙 √𝟐 𝟏 𝐬𝐢𝐧−𝟏 +𝑪 = √𝟐 √𝟑 𝑑𝑥

∫ 2+ 7𝑥2

7.

Solution: = ∫

𝑑𝑥

1

𝑑𝑢

= 7 ∫ 𝑎 2 + 𝑢2

3 7( +𝑥 2 ) 2

3

√3

Let a2 = 2 ; a = ; u = x; du = dx √2 1 1 𝑢 = arctan 𝑎 7 𝑎 1 √3 𝑥 +𝐶 = tan−1 7 √2 √3 =

√𝟑 𝟕√𝟐

8.

√2 +c arctan √𝟐𝒙 √𝟑

∫

𝑑𝑥

𝑥√𝑥 2 −4

Solution: 𝐿𝑒𝑡:

𝑢 = 𝑥,

𝑑𝑢 = 𝑑𝑥, 𝑎 = 2

3

Integral Calculus Integration Leading to Inverse Trigonometric Functions

= √ 𝑥 2 − 4 = √ 𝑢 2 − 𝑎2 1 𝑑𝑢 𝑢 1 = ∫ = sec−1 𝑎 + 𝐶 𝑎 𝑢√𝑢2 − 1 𝑎 𝒙 𝟏 −𝟏 = 𝐬𝐞𝐜 ( ) + 𝑪 𝟐 𝟐

4...