Further Calculus IV - Cheat Sheet - Differentiating Inverse Trigonometric Functions and Using Them in Integration

Q: When differentiating sin−1(x), why must |x|<1 be used?

The inverse sine is defined with a restricted input range because the original range of sin is [−1,1]. At x=±1 the gradient of y=sin−1(x) becomes infinite, so the derivative is undefined there, leading to the restriction |x|<1.

Q: How are derivatives of inverse trigonometric functions used in integration questions?

Integration problems are arranged into integrands matching standard forms whose antiderivatives involve sin−1(x) or tan−1(x). After rewriting the integrand and applying the derivative identities through substitution, the result includes the constant of integration.

Oct 14, 2026

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Cheat sheet summarizing key A Level calculus techniques for inverse trigonometric functions. Covers how to differentiate inverse sine, cosine, and tangent using implicit differentiation and domain restrictions such as |x|<1. Includes derivative forms used in formula book contexts, and shows how these results support integration problems via inverse-trig substitutions. Provides example workflows for both differentiation and integration, including identifying domains, applying chain/product/quotient rules, and rewriting integrands into standard inverse-trig forms.

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Differentiating inverse trigonometric functions

Implicit differentiation and domain restrictions

Derivatives of arcsin, arccos, arctan

Using chain, product, and quotient rules

Using inverse trigonometric functions in integration

Standard integral forms and constant of integration

Substitution and algebraic rearrangement

Worked examples and proof outlines

Further Calculus IV Cheat Sheet A Level Further AQA Core Maths: Differentiating Inverse Trigonometric Functions (A Level Only) To find the derivatives of inverse trigonometric functions, implicit differentiation is used. In order for a function to have an inverse, it must have a 1-to-1 mapping between the domain and the range. For example consider sin(􀝔), its range is [−1, 1] and so its inverse, sin−1(􀝔), is defined for 􀝔 ∈ [−1, 1] . The graph of y = sin−1(􀝔) is shown to the right. Although this function is defined at 􀝔 = ±1, the gradient at these points is infinite, and therefore undefined. Thus, the resitrction |􀝔| < 1 is taken when finding the derivative of this function. Example 1: Given |􀝔| < 1, find 􀯗􀯗􀯬􀯫 for 􀝕 = sin−1(􀝔) . The derivatives of cos−1(􀝔) and tan−1(􀝔) are found in a similar way, giving: 􀝀 􀝀􀝔 (cos−1(􀝔)) = − 1 √1 − 􀝔 2 , |􀝔| < 1 􀝀 􀝀􀝔 (tan−1(􀝔)) = 1~~ ~~􀝔~~ ~~2 . These are given in the formula book. The derivative of tan−1(􀝔) is defined for all 􀝔, since the range of tan(􀝔) is infinite. These results can also be used within examples which involve the product, chain and quotient rules. Example 2: Find the derivative of 􀝕 = 􀝁 6􀯫 arctan(2􀝔) . To the right is the graph of 􀝕 = tan−1(􀝔) . The domain of this function is (−∞, ∞) since tan(􀝔) takes on an infinite number of values, unlike sin (􀝔) and cos(􀝔) . The range of this function is ( − 􀰗2 ~~ ~~ , 􀰗 2 ), with the graph approaching each limit asymptotically. This is due to tan(􀝔) being undefined at ± 􀰗2 . Using Inverse Trigonometric Functions in Integration (A Level Only) The derivatives of sin−1(􀝔) and tan−1(􀝔) are used to evaluate integrals of the form ∫ √􀯔~~2~~1−􀯫~~2~~ 􀝀􀝔 and ∫ 􀯔2􀯫2 􀝀􀝔 respectively. The derivative for cos−1(􀝔) does not offer any additional help here as it is the negative of the derivative of sin−1(􀝔) . The general results that are used are as follows: ∫ √􀜽2~~ ~~~~ ~~􀝔~~ ~~2~~ ~~ 􀝀􀝔 = sin−1 (􀝔􀜽 ) + 􀜿, |􀝔| > 􀜽 ∫ 1 􀜽2 + 􀝔 2 􀝀􀝔 = 1 􀜽 tan−1 (􀝔􀜽 ) + 􀜿 where 􀜿 is a constant of integration. These are given in the formula booklet. Alongside using these identities, question types may include proving the above results. Example 3: Use the substitution 􀝔 = 􀜽 tan(􀝑) to prove the result ∫ 􀯔2􀯫2~~ ~~ ~~ ~~ ~~ ~~ 􀝀􀝔 = 1 􀯔 tan−1 (􀯫􀯔 ) + 􀜿 . Note: the result for ∫ √􀯔~~2~~􀯫~~2 ~~ 􀝀􀝔 is similarly proven. To complete questions using these results, the integrand may need to be rearranged into the desired form Example 4: Find- ∫ √−􀝔~~2 ~~−12􀝔~~ ~~+~~ ~~35 􀝀􀝔 | Apply the sine function to find 􀝔 in terms of 􀝕 . This is now differentiable via implicit differentiation. | 􀝕 = sin−1(􀝔) ⇒ sin(􀝕) = 􀝔 . | | | --- | --- | --- | | Differentiate this equation with respect to 􀝔 . The aim is to have an expression for 􀯗􀯗􀯬􀯫 in terms of 􀝔 . | Using the chain rule, cos(􀝕) 􀯗􀯗􀯬􀯫 = 1 | | | | ⇒ | 􀝀􀝕 1 ~~ ~~ = ~~ ~~􀝀􀝔 cos(􀝕) . | | | | | --- | --- | | Use trigonometric identities to express cos(􀝕) in terms of 􀝔 . The range of sin−1(􀝔) is − 􀰗2 ≤ 􀝕 ≤ 􀰗2, and since cos(􀝕) is non-negative in this range, it is justified to take just the positive square root here. Looking at the graph of sin−1(􀝔) also confirms that the gradient is only ever positive. | cos2 (􀝕) = 1 − sin2 (􀝕)
⇒ cos(􀝕) = √1 − sin2 (􀝕) .
It is given that sin(􀝕) = 􀝔, and so this expression becomes
􀝀􀝕 1 1
~~ ~~ = ~~ ~~ = ~~ ~~
􀝀􀝔 √1 − sin2 (􀝕) √1 − 􀝔 2. | | There are two functions multiplied together here, so the product rule must be used. Note that arctan (􀝔)
is different notation for tan−1(􀝔). The chain rule is used to find the derivative of arctan(2􀝔) . 2􀝔 is used in place of 􀝔 in the result for its derivative. | Product rule: 􀯗􀯗􀯫 (􀝑􀝒) = 􀝑′􀝒 + 􀝒′􀝑 . 􀝑 = 􀝁 6􀯫 , 􀝑′ = 6􀝁 6􀯫 .
Using the chain rule (􀝂(􀝃 (􀝔))′ = 􀝂 ′(􀝃 (􀝔))􀝃 ′(􀝔) with 􀝂(􀝔) = arctan(􀝔), 􀝃 (􀝔) = 2􀝔:
􀝒 = arctan(2􀝔) , 􀝒 ′ = 11+∙~~ ~~((22􀝔􀝔))′2 = 1~~ ~~+24􀝔2 ∴ 􀝑′􀝒 + 􀝒′􀝑 = 6􀝁6􀯫 arctan(2􀝔) + 1~~ ~~􀝁~~ ~~64􀯫􀝔2 ∴ 􀝀􀝀􀝔 (􀝁6􀯫 arctan(2

FAQs

Why is implicit differentiation required for inverse trigonometric functions?

Inverse trig derivatives are derived by implicitly differentiating the relationship between the function and its inverse, then isolating the derivative. This ensures the derivative respects the inverse mapping between domain and range.

When differentiating sin−1(x), why must |x|<1 be used?

The inverse sine is defined with a restricted input range because the original range of sin is [−1,1]. At x=±1 the gradient of y=sin−1(x) becomes infinite, so the derivative is undefined there, leading to the restriction |x|<1.

How are derivatives of inverse trigonometric functions used in integration questions?

Integration problems are arranged into integrands matching standard forms whose antiderivatives involve sin−1(x) or tan−1(x). After rewriting the integrand and applying the derivative identities through substitution, the result includes the constant of integration.

Oct 14, 2026

Upload by Rainbow Cat