3.2 Answer

3.2

(a) Let $t = 2 - x$ . Then $dt = \frac{dt}{dx}dx = -dx$ . Thus, the given integral is a function of $t$ and $dt$ . Thus,

$\displaystyle \int e^{2-x} dx = - \int e^{t} dt = - e^{t} + c \underbrace{=}_{t = 2-x} {e^{2 - x}} + c$

(b)

$Let t = 1 - x$ . Then $dt = -dx$ . Thu, we can express as a function of $t$ , $\sec^{2}{t}$ . Thus,

$\displaystyle \int \sec^{2}{(1 - x)}dx = - \int \sec^{2}{t} dt = -\tan (1 - x) + c$

(c)

Let $t = 1 - x^2$ . Then $dt = -2xdx$ and

$\displaystyle \int \frac{x}{\sqrt{1 - x^2}} dx = -\frac{1}{2} \int \frac{dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-\frac{1}{2}} dt = -{\sqrt{1 - {x^2}}} + c$

(d)

Let $t = \cos{x}$ . Then $dt = -\sin{x}dx$ and

$\displaystyle \int \frac{\sin{x}}{\cos^{2}{x}} dx = - \int \frac{dt}{t^{2}} = -\int t^{-2}dt = t^{-1} + c = \sec(x) + c$

(e)

Let $t = \frac{1}{x}$ . Then $dt = - \frac{1}{x^2} dx$ and

$\displaystyle \int \frac{e^{1/x}}{x^2} dx = -\int e^{t}dt = -e^{t} + c = - e^{\frac{1}{x}} + c$

(f)

Let $t = 3\tan{\theta} + 1$ . Then $dt = 3\sec^{2}{\theta} d \theta$ and

$\displaystyle \int \frac{sec^{2}{\theta}}{3\tan{\theta} + 1} d \theta = \frac{1... ...t}} = \frac{2}{3}\sqrt{t} + c = \frac{2\,{\sqrt{1 + 3\,\tan (\theta)}}}{3} + c$

(g)

$\displaystyle \int \frac{1 + \cos{2x}}{\sin^{2}{x}}dx$	$\displaystyle =$	$\displaystyle \int \frac{\cos^{2}{x}}{\sin^{2}{x}}dx$
	$\displaystyle =$	$\displaystyle 2\int \frac{1 - \sin^{2}{x}}{\sin^{2}{x}}dx = 2\int({\rm cosec}^{2}{x} - 1)dx$
	$\displaystyle =$	$\displaystyle 2(-\cot{x} - x) + c$