Answer to Question #247643 in Calculus for JaytheCreator

Question

Question 1 Find (i) ∫ 𝑑𝑥 𝑥(ln 𝑥)2, (ii) ∫ 𝑥√𝑥 + 1 𝑑𝑥, (iii) ∫ 𝑥(𝑥 2 + 3) 4𝑑𝑥 1 0 , (iv) cos2 𝑥 sin3 𝑥 𝑑𝑥. Try 𝑡 = sin x

Answer 1

$\displaystyle 2\int x\ln x \\ \text{Using integration by parts, we evaluate the above integral}\\ \text{Let $u = \ln x, dv = x$ }\implies du = \frac{1}{x}, v = \frac{x^2}{2}\\ \int udv = uv -\int vdu\\ \implies 2\int x\ln x = 2(\frac{x^2}{2}\ln x - \frac{x^2}{4}+c)\\ x^2\ln x - \frac{x^2}{2}+c\\ \int (x\sqrt{x}+1)dx = \int (x^{\frac{3}{2}}+1)dx\\ = \frac{2x^{\frac{5}{2}}}{5}+x+c\\ \int x(x^2+3)4dx= \int (4x^3 +12x)dx\\ x^4+6x^2+c\\ \int \cos 2x \sin 3x dx\\ \sin A \cos B = \frac{1}{2}(\sin 5x + \sin x)dx\\ \therefore \int \cos 2x \sin 3x dx = \frac{1}{2}\int \sin 5xdx +\frac{1}{2} \int \sin xdx\\ =-\frac{1}{10}\cos5x-\frac{1}{2}\cos x+c$