Calculus 2 - Integration by Trigonometric Substitution
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Course
Calculus
Institution
University Of East London (UEL)
This document will allow university Math students to practice integration by trigonometric substitution. All the questions are similar in style and full worked solutions have been provided. This will allow you to understand how a certain question in this document is answered and then you will be ab...
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.
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