Questions on calculus with inverse trigonometric functions

Question 1

Write down the derivatives of arcsin x and arctan x.

Accepted Answer

d/dx (arcsin x) = 1/√(1 − x²) and d/dx (arctan x) = 1/(1 + x²). B1 B1. Neither derivative contains a trigonometric function, so read backwards the two results give standard integrals.

Question 2

Starting from y = arctan x, so that tan y = x, derive the derivative of arctan x.

Accepted 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. Without the identity the answer is left as 1/sec²y rather than in terms of x.

Question 3

Write down ∫1/√(16 − x²) dx and ∫1/(16 + x²) dx.

Accepted Answer

With a = 4 in both cases: ∫1/√(16 − x²) dx = arcsin(x/4) + c and ∫1/(16 + x²) dx = (1/4)arctan(x/4) + c. B1 B1. The square root under the first pattern and the sum in the second are what tell them apart; the arctan form carries the extra factor 1/a and the arcsin form does not.

Question 4

Given that arcsin x + arccos x = π/2 for −1 ≤ x ≤ 1, find the derivative of arccos x.

Accepted Answer

Differentiating both sides with respect to x: d/dx(arcsin x) + d/dx(arccos x) = 0, since π/2 is constant. M1 for differentiating the identity. So d/dx(arccos x) = −d/dx(arcsin x) = −1/√(1 − x²). A1 for the printed result. The two derivatives are negatives of each other on |x| < 1 because the identity holds at every such x, not only at one point: a relationship true throughout an interval differentiates to another relationship true throughout it. B1 for that reasoning.

Question 5

Find the gradient of the curve y = arcsin 2x at the point where x = 1/4.

Accepted Answer

Apply the chain rule: d/dx(arcsin 2x) = 2/√(1 − 4x²). The factor 2 is the derivative of the inner function.
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.

Question 6

Evaluate ∫ 1/(4 + x²) dx from 0 to 2, giving an exact answer.

Accepted 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.

Question 7

Evaluate ∫ 1/√(9 − x²) dx from 0 to 3/2, giving an exact answer.

Accepted 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.

Question 8

Find ∫ (x² + 7)/(x² + 4) dx.

Accepted Answer

The numerator's degree matches the denominator's, so the fraction is improper and must be divided down before anything else. Force the denominator out of the numerator: x² + 7 = (x² + 4) + 3, so the integrand is 1 + 3/(x² + 4). M1 for the division, A1 for that form. The constant integrates to x, and 3/(x² + 4) is the arctan pattern with a = 2, carrying its 1/a factor: 3 × (1/2) arctan(x/2). M1 for the pattern with a = 2. The integral is x + (3/2) arctan(x/2) + c. A1. Setting the improper fraction equal to partial fractions without dividing first produces contradictory coefficients, because one side is proper and the other is not.

Question 9

Starting from y = arcsin x, so that sin y = x, derive the derivative of arcsin x, justifying the sign you take for cos y.

Accepted Answer

Differentiate implicitly: cos y (dy/dx) = 1. M1 for the implicit differentiation. Since sin²y + cos²y = 1, cos y = ±√(1 − sin²y) = ±√(1 − x²), and on the principal branch −π/2 ≤ y ≤ π/2, cos y ≥ 0, so the positive root is the one that applies. B1 for taking the positive square root, with the branch as the reason. M1 for replacing cos y by √(1 − x²). So dy/dx = 1/√(1 − x²). A1 for the printed result. The sign has to be argued: outside the principal branch cos y can be negative.

Question 10

Evaluate ∫ 1/(x² − 4x + 13) dx from 0 to 4, giving an exact answer.

Accepted Answer

Complete the square on the denominator: x² − 4x + 13 = (x − 2)² + 9. M1 for completing the square. This matches the arctan pattern with a = 3 and x replaced by x − 2, so the integral is (1/3)arctan((x − 2)/3). A1 for the antiderivative. At the limits: (1/3)[arctan(2/3) − arctan(−2/3)] = (2/3)arctan(2/3), using arctan(−2/3) = −arctan(2/3). M1 for substituting the limits. The exact value is (2/3)arctan(2/3), about 0.392. A1 for the exact value. A quadratic with no x-term to split off needs only the completed square, not the numerator work the next pattern requires.

Question 11

Use the substitution x = 3 sin θ to show that ∫ 1/√(9 − x²) dx = arcsin(x/3) + c, confirming the standard result directly.

Accepted Answer

With x = 3 sin θ, dx = 3 cos θ dθ, and 9 − x² = 9 − 9 sin²θ = 9 cos²θ, so √(9 − x²) = 3 cos θ, taking the positive root since θ lies in [−π/2, π/2] there. M1 for the substitution with dx and the simplified root. The integral becomes ∫ 3 cos θ/(3 cos θ) dθ = ∫ 1 dθ = θ + c. A1 for the simplification to ∫1 dθ. Since x = 3 sin θ gives θ = arcsin(x/3), the result is arcsin(x/3) + c, exactly the standard form quoted for a = 3. M1 for θ = arcsin(x/3), A1 for the printed result.

Question 12

Show that ∫ (2x + 4)/(x² + 2x + 5) dx from −1 to 1 = ln 2 + π/4.

Accepted 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.
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.

Question 13

Explain why the substitution x = 3 sin θ, rather than x = 3 tan θ, suits the integral ∫ √(9 − x²) dx from 0 to 3, and hence evaluate it exactly.

Accepted Answer

The root fits neither read-off pattern, so substitute so that a Pythagorean identity collapses the quadratic: with x = 3 sin θ, 9 − x² = 9 − 9 sin²θ = 9 cos²θ, and the square root disappears; x = 3 tan θ would give 9 + 9 tan²θ = 9 sec²θ, which matches 9 + x², not 9 − x². B1 for the choice argued through the identity. Then dx = 3 cos θ dθ, and the limits x = 0 and x = 3 become θ = 0 and θ = π/2. M1 for the substitution carried through with dx and the limits. √(9 − x²) = 3 cos θ, taking the positive root since cos θ ≥ 0 across [0, π/2], so the integral is ∫ 9 cos²θ dθ = ∫ (9/2)(1 + cos 2θ) dθ. A1 for that form. Evaluating: (9/2)[θ + (sin 2θ)/2] from 0 to π/2 = (9/2)(π/2) = 9π/4. M1 for the double-angle step, A1 for the exact value. The answer is a quarter of the circle x² + y² = 9, area πr²/4 with r = 3, so the substitution has recovered the geometry; the question asks for an exact value, so leave it as 9π/4.

Question 14

Find ∫ (x + 7)/((x − 1)(x² + 9)) dx.

Accepted Answer

The quadratic factor needs a linear numerator, so write (x + 7)/((x − 1)(x² + 9)) = A/(x − 1) + (Bx + C)/(x² + 9). M1 for setting up the decomposition with a linear numerator over the quadratic factor. Multiplying up, x + 7 = A(x² + 9) + (Bx + C)(x − 1). Putting x = 1 gives 10A = 8, so A = 4/5; comparing x² terms gives B = −4/5, and comparing constant terms gives C = 9A − 7 = 1/5. M1 for the substitution and the coefficient comparison, A1 for A = 4/5, B = −4/5, C = 1/5. The second fraction splits again, into −(4/5)x/(x² + 9), which integrates by recognition since its numerator is (up to a constant) the derivative of x² + 9, and (1/5)/(x² + 9), the arctan pattern with a = 3. M1 for splitting the second fraction into these two pieces. Integrating: (4/5)ln|x − 1| − (2/5)ln(x² + 9) + (1/15)arctan(x/3) + c. A1 for the logarithm terms, A1 for the arctan term. Every irreducible quadratic factor carries a linear numerator Bx + C, never a constant alone, because a constant numerator cannot supply the x-term needed to match the numerator once the equation is multiplied out.

Question 15

By substituting x = a tan θ, where a > 0, derive the general result ∫ 1/(a² + x²) dx = (1/a)arctan(x/a) + c, valid for all real x.

Accepted Answer

With x = a tan θ, dx = a sec²θ dθ, and a² + x² = a² + a²tan²θ = a²sec²θ, using 1 + tan²θ = sec²θ. M1 for the substitution with dx. A1 for a² + x² = a²sec²θ. The integral becomes ∫ a sec²θ/(a²sec²θ) dθ = ∫ (1/a) dθ, since the sec²θ factors cancel completely. M1 for the simplification to a constant integrand. Integrating, θ/a + c, and since x = a tan θ gives θ = arctan(x/a) for every real x, without any restriction on θ's range being needed, the result is (1/a)arctan(x/a) + c. A1 for θ = arctan(x/a), A1 for the printed result.