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Calculus with inverse trigonometric functions questions
Differentiating arcsin, arccos and arctan produces algebraic fractions rather than trigonometric ones. Read in reverse, those results integrate whole families of expressions containing a² − x² or a² + x², which have no obvious antiderivative otherwise.
6 original questions · 22 marks · the calculus with inverse trigonometric functions notes · Further calculus
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Write down the derivatives of arcsin x and arctan x.
Worked answer
d/dx (arcsin x) = 1/√(1 − x²) and d/dx (arctan x) = 1/(1 + x²). B1 B1. Neither derivative contains a trigonometric function, which is what makes the two results so useful read backwards, as standard integrals.Starting from y = arctan x, so that tan y = x, derive the derivative of arctan x.
Worked answer
Differentiate implicitly to get sec²y (dy/dx) = 1. Since sec²y = 1 + tan²y = 1 + x², dy/dx = 1/(1 + x²). M1 for the implicit differentiation, M1 for replacing sec²y by 1 + x², A1 for the printed result. The identity is the whole point of the derivation; it turns the trigonometry in the denominator into algebra in x, and without it the answer is left as 1/sec²y and scores no accuracy mark.Find the gradient of the curve y = arcsin 2x at the point where x = 1/4.
Worked answer
The chain rule brings the inner derivative to the front, so dy/dx = 2/√(1 − 4x²). Dropping that factor of 2 is the single most common slip here, and it loses the method mark.
At x = 1/4: 2/√(1 − 1/4) = 2/√(3/4) = 4/√3, about 2.31. M1 A1 for the derivative, A1 for the gradient.Evaluate ∫ 1/(4 + x²) dx from 0 to 2, giving an exact answer.
Worked answer
Match the form a² + x² with a = 2, so the integral is (1/2) arctan(x/2). The 1/a in front belongs to the arctan pattern; omitting it doubles the final answer.
At the limits: (1/2)(arctan 1 − arctan 0) = (1/2)(π/4) = π/8. M1 for the arctan form, A1 for the antiderivative, M1 for substituting the limits, A1 for the exact value.Evaluate ∫ 1/√(9 − x²) dx from 0 to 3/2, giving an exact answer.
Worked answer
Match the form a² − x² under the root with a = 3, so the integral is arcsin(x/3), with no factor in front. The arcsin pattern and the arctan pattern differ in exactly that respect.
At the limits: arcsin(1/2) − arcsin 0 = π/6. M1 for the arcsin form, A1 for the antiderivative, M1 for substituting the limits, A1 for the exact value.Show that ∫ (2x + 4)/(x² + 2x + 5) dx from −1 to 1 = ln 2 + π/4.
Worked answer
Neither standard form fits as it stands, so split the numerator into the derivative of the denominator plus a constant: 2x + 4 = (2x + 2) + 2. That split is the first method mark, and an attempt at a single arctan or a single logarithm gets nowhere.
The first piece integrates by recognition, since its numerator is exactly the derivative of the denominator, giving ln(x² + 2x + 5).
For the second piece, complete the square: x² + 2x + 5 = (x + 1)² + 4, so ∫2/((x + 1)² + 4) dx = 2 × (1/2) arctan((x + 1)/2) = arctan((x + 1)/2).
Evaluating from −1 to 1: the logarithm gives ln 8 − ln 4 = ln 2, and the arctan gives arctan 1 − arctan 0 = π/4. The total is ln 2 + π/4, about 1.479, as required.
M1 A1 for the split of the numerator, A1 for the logarithm term, M1 for completing the square, A1 for the arctan term, A1 for the printed result. Any quadratic denominator with no real roots yields this same pair of pieces, a logarithm and an arctan.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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