Question

evaluate the integral using any appropri\\(\int (\csc x + \tan x)^2 dx\\)

Explanation:

Step1: Expand the squared integrand

Use $(a+b)^2=a^2+2ab+b^2$
$$(\csc x + \tan x)^2 = \csc^2 x + 2\csc x \tan x + \tan^2 x$$

Step2: Rewrite trigonometric terms

Express in sine/cosine, use $\tan^2 x=\sec^2 x-1$
$$\csc^2 x + 2\cdot\frac{1}{\sin x}\cdot\frac{\sin x}{\cos x} + \sec^2 x - 1 = \csc^2 x + 2\sec x + \sec^2 x - 1$$

Step3: Split the integral into parts

Integrate term by term
$$\int \csc^2 x dx + 2\int \sec x dx + \int \sec^2 x dx - \int 1 dx$$

Step4: Apply standard integrals

Use known antiderivatives

• $\int \csc^2 x dx = -\cot x$, $\int \sec x dx = \ln|\sec x + \tan x|$, $\int \sec^2 x dx = \tan x$, $\int 1 dx = x$

$$-\cot x + 2\ln|\sec x + \tan x| + \tan x - x + C$$

Answer:

$\tan x - \cot x - x + 2\ln|\sec x + \tan x| + C$ (where $C$ is the constant of integration)