Calculus 2 - Integration by Trigonometric Substitution
Case 1:
∫ 𝑎² − 𝑥² 𝑑𝑥
Use substitution 𝑥 = 𝑎𝑠𝑖𝑛𝜃
𝑑𝑥 = 𝑎𝑐𝑜𝑠𝜃𝑑𝜃
Case 2:
∫ 𝑥² − 𝑎² 𝑑𝑥
Use Substitution 𝑥 = 𝑎𝑠𝑒𝑐𝜃
𝑑𝑥 = 𝑎𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
Case 3:
∫ 𝑥² + 𝑎² 𝑑𝑥 OR ∫ 𝑎² + 𝑥² 𝑑𝑥
Use Substitution 𝑥 = 𝑎𝑡𝑎𝑛𝜃
𝑑𝑥 = 𝑎𝑠𝑒𝑐²𝜃𝑑𝜃
, 1) ∫ 49 − 𝑥² 𝑑𝑥
Case 1:
𝑎² = 49
𝑎 = 49
𝑎 = 7
𝑥 = 7𝑠𝑖𝑛𝜃
𝑑𝑥 = 7𝑐𝑜𝑠𝜃𝑑𝜃
Substitute 𝑥 = 7𝑠𝑖𝑛𝜃 and 𝑑𝑥 = 7𝑐𝑜𝑠𝜃𝑑𝜃 into the original integral
∫ 49 − (7𝑠𝑖𝑛𝜃)² 7𝑐𝑜𝑠𝜃 𝑑𝜃
This integral is now in trigonometric terms
∫ 49 − 49𝑠𝑖𝑛²𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫ 49(1 − 𝑠𝑖𝑛²𝜃) 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫ 49𝑐𝑜𝑠²𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫7𝑐𝑜𝑠𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫49𝑐𝑜𝑠²𝜃 𝑑𝜃
49∫𝑐𝑜𝑠²𝜃 𝑑𝜃
Power reduction formula for cos²𝜃:
1 1
𝑐𝑜𝑠²𝜃 = 2
+ 2
𝑐𝑜𝑠2𝜃
1 1
49∫ 2 + 2
𝑐𝑜𝑠2𝜃 𝑑𝜃
1 1
49 ( 2 𝜃 + 4
𝑠𝑖𝑛2𝜃)
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
,The expression in the green box is the answer however, we need to convert this
answer to an expression in terms of 𝑥 since the original question was in terms of 𝑥.
We will add the + C at the very end in the final answer to avoid any confusion.
Sin Double Angle Formula: 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 49
2
𝜃 + 4
(2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)
49 49
2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑥 = 7𝑠𝑖𝑛𝜃 → This is the first substitution we used at the beginning of the question.
𝑥 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑠𝑖𝑛𝜃 = 7
→ 𝑠𝑖𝑛𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑥
49 − 𝑥²
𝑐² = 𝑎² + 𝑏²
7² = 𝑥² + 𝑏²
49 = 𝑥² + 𝑏²
𝑏² = 49 − 𝑥²
𝑏 = 49 − 𝑥²
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
49 49
= 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑥
𝑠𝑖𝑛𝜃 = 7
𝑥
𝜃 = 𝑠𝑖𝑛⁻¹( 7 )
, 49−𝑥²
From Triangle: 𝑐𝑜𝑠𝜃 = 7
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
49 49
= 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 49
Substitute 𝜃, 𝑐𝑜𝑠𝜃 and 𝑠𝑖𝑛𝜃 into 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 𝑥 49 𝑥 49−𝑥²
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
( 7 )( 7
)
Simplifying the expression above:
49 𝑥 1
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
𝑥 49 − 𝑥²
FINAL ANSWER:
49 𝑥 1
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
𝑥 49 − 𝑥² + 𝐶
Case 1:
∫ 𝑎² − 𝑥² 𝑑𝑥
Use substitution 𝑥 = 𝑎𝑠𝑖𝑛𝜃
𝑑𝑥 = 𝑎𝑐𝑜𝑠𝜃𝑑𝜃
Case 2:
∫ 𝑥² − 𝑎² 𝑑𝑥
Use Substitution 𝑥 = 𝑎𝑠𝑒𝑐𝜃
𝑑𝑥 = 𝑎𝑠𝑒𝑐𝜃𝑡𝑎𝑛𝜃𝑑𝜃
Case 3:
∫ 𝑥² + 𝑎² 𝑑𝑥 OR ∫ 𝑎² + 𝑥² 𝑑𝑥
Use Substitution 𝑥 = 𝑎𝑡𝑎𝑛𝜃
𝑑𝑥 = 𝑎𝑠𝑒𝑐²𝜃𝑑𝜃
, 1) ∫ 49 − 𝑥² 𝑑𝑥
Case 1:
𝑎² = 49
𝑎 = 49
𝑎 = 7
𝑥 = 7𝑠𝑖𝑛𝜃
𝑑𝑥 = 7𝑐𝑜𝑠𝜃𝑑𝜃
Substitute 𝑥 = 7𝑠𝑖𝑛𝜃 and 𝑑𝑥 = 7𝑐𝑜𝑠𝜃𝑑𝜃 into the original integral
∫ 49 − (7𝑠𝑖𝑛𝜃)² 7𝑐𝑜𝑠𝜃 𝑑𝜃
This integral is now in trigonometric terms
∫ 49 − 49𝑠𝑖𝑛²𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫ 49(1 − 𝑠𝑖𝑛²𝜃) 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫ 49𝑐𝑜𝑠²𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫7𝑐𝑜𝑠𝜃 7𝑐𝑜𝑠𝜃 𝑑𝜃
∫49𝑐𝑜𝑠²𝜃 𝑑𝜃
49∫𝑐𝑜𝑠²𝜃 𝑑𝜃
Power reduction formula for cos²𝜃:
1 1
𝑐𝑜𝑠²𝜃 = 2
+ 2
𝑐𝑜𝑠2𝜃
1 1
49∫ 2 + 2
𝑐𝑜𝑠2𝜃 𝑑𝜃
1 1
49 ( 2 𝜃 + 4
𝑠𝑖𝑛2𝜃)
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
,The expression in the green box is the answer however, we need to convert this
answer to an expression in terms of 𝑥 since the original question was in terms of 𝑥.
We will add the + C at the very end in the final answer to avoid any confusion.
Sin Double Angle Formula: 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 49
2
𝜃 + 4
(2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)
49 49
2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑥 = 7𝑠𝑖𝑛𝜃 → This is the first substitution we used at the beginning of the question.
𝑥 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑠𝑖𝑛𝜃 = 7
→ 𝑠𝑖𝑛𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑥
49 − 𝑥²
𝑐² = 𝑎² + 𝑏²
7² = 𝑥² + 𝑏²
49 = 𝑥² + 𝑏²
𝑏² = 49 − 𝑥²
𝑏 = 49 − 𝑥²
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
49 49
= 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑥
𝑠𝑖𝑛𝜃 = 7
𝑥
𝜃 = 𝑠𝑖𝑛⁻¹( 7 )
, 49−𝑥²
From Triangle: 𝑐𝑜𝑠𝜃 = 7
49 49
2
𝜃 + 4
𝑠𝑖𝑛2𝜃
49 49
= 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 49
Substitute 𝜃, 𝑐𝑜𝑠𝜃 and 𝑠𝑖𝑛𝜃 into 2
𝜃 + 2
𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
49 𝑥 49 𝑥 49−𝑥²
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
( 7 )( 7
)
Simplifying the expression above:
49 𝑥 1
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
𝑥 49 − 𝑥²
FINAL ANSWER:
49 𝑥 1
2
𝑠𝑖𝑛⁻¹( 7 ) + 2
𝑥 49 − 𝑥² + 𝐶
