Answer to Question #202791 in Calculus for Ayush

Question

Integrate the following functions w.r.t. x:

1. 1/(2x+1)^3/2

2. Sin(2x+3)

3. Cosec(4x)

4. 1/(1-9x²)^1/2

Answer 1

$\displaystyle 1)\\ \int \frac{1}{(2x+1)^{\frac{3}{2}}}\,\,\mathrm{d}x = -2(2x+1)^{\frac{-1}{2}} \times \frac{1}{2} + C = -(2x+1)^{\frac{-1}{2}} + C\\ 2)\\ \int \sin(2x + 3)\,\,\mathrm{d}x = -\frac{\cos(2x + 3)}{2} + C\\ 3)\\ \begin{aligned} \int \cosec(4x)\,\mathrm{d}x &= \int \cosec(4x)\frac{\cosec(4x) + \cot(4x)}{\cosec(4x) + \cot(4x)}\,\mathrm{d}x \\&= \int\frac{\cosec^2(4x) + \cosec(4x)\cot(4x)}{\cosec(4x) + \cot(4x)}\,\mathrm{d}x \\&= -\int\frac{\mathrm{d}\left(\cosec(4x) + \cot(4x)\right)}{4(\cosec(4x) + \cot(4x))} \\&= -\frac{1}{4}\ln\left|\cosec(4x) + \cot(4x)\right| + C \end{aligned} \\ 4)\\ \begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{1 - 9x^2}} &= \frac{1}{\sqrt{9}}\int \frac{\mathrm{d}x}{\sqrt{\frac{1}{9} - x^2}} \\&= \frac{1}{3}\int \frac{\mathrm{d}x}{\sqrt{\frac{1}{9} - x^2}} \\&= \frac{1}{3}\arcsin\left(\frac{x}{1/3}\right) + C \\&= \frac{1}{3}\arcsin\left(3x\right) + C \end{aligned}$