Additional Mathematics 0606 / Calculus — Integration / 1.251.25: Logarithmic and trigonometric integration0606/11/O/N/20 — Question 9 · 11 marksMark as done · Save for later · Show all solutions(a)Given that ∫1a(1x−12x+3)dx=ln3\displaystyle\int_{1}^{a}\left(\frac{1}{x} - \frac{1}{2x + 3}\right)\mathrm{d}x = \ln 3∫1a(x1−2x+31)dx=ln3, where a>0a > 0a>0, find the exact value of aaa, giving your answer in simplest surd form.[6]▸ Answer(b)Find the exact value of ∫0π/3(sin(2x+π3)−1+cos2x)dx\displaystyle\int_{0}^{\pi/3}\left(\sin\left(2x + \frac{\pi}{3}\right) - 1 + \cos 2x\right)\mathrm{d}x∫0π/3(sin(2x+3π)−1+cos2x)dx.[5]▸ Answer▸ Official mark scheme← 3.88: Area between a cubic curve and a horizontal line1.86: Partial fractions and shaded area →