Use the substitution u = cosx to show that ∫(from 0 to π/2) sin 2x e^(2 cos x) dx = ∫(from 1 to -1) 2u e^(2u) du.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] du/dx = -sin x; SOI
2. [M1] Use double angle formula and substitute for x and dx throughout the integral; All x's must be removed, can be coefficient errors provided 2 seen in working.
3. [A1] Obtain ± ∫ 2ue^(2u) du; Limits may be omitted, or left as 0 and π, during the change of variable stage.
4. [A1] Justify new limits and obtain ∫(from 1 to -1) 2ue^(2u) du from correct working; AG Must see x = 0, u = 1 and x = π/2, u = -1. Inequalities alone e.g. 0 ≤ x ≤ π/2 and 1 ≤ u ≤ -1 or -1 < u < 1 for limits are insufficient A0 If sign in expression and order of limits incorrect then A0. If negative sign is present in the integrand then this can be removed and limits introduced in correct order in a single step.
5. [M1*] Commence integration and reach aue^(2u) + b∫e^(2u)du, where ab≠0, b< 0; Condone dx.
6. [A1] Complete integration and obtain ue^(2u) – (1/2)e^(2u); OE Allow (2u - 1)e^(2u) / 2.
7. [DM1] Use correct limits correctly in cue^(2u) + d e^(2u) having integrated twice or in c cos x e^(2 cos x) + d e^(2 cos x); 1 and -1 for u, 0 and π/2 for x e.g. ce² + de² - (- ce⁻² + de⁻²). Not decimals. Allow one sign error at most in going from cue^(2u) + d e^(2u) or c cos x e^(2 cos x) + d e^(2 cos x) to ce² + de² – (- ce⁻² + de⁻²). [e² - (1/2)e² - (-e⁻² – (1/2)e⁻²)] Complete reversal of sign by converting back to cos x and not making x = 0 upper limit is DM0 A0.
8. [A1] Obtain (e²+3)/(2e²); ISW Or equivalent 2-term expression e.g. (e^(4)+3)/(2e²) or e²/2 + 3/(2e²).