Question

Integration of sec square x cosecx square x

Answer 1

tan x - cot x + C

Answer 2

To integrate $\sec^2 x \operatorname{cosec}^2 x$, we can use trigonometric identities.

The given integral is: $I = \int \sec^2 x \operatorname{cosec}^2 x \, dx$

Step 1: Rewrite $\sec^2 x$ and $\operatorname{cosec}^2 x$ in terms of $\sin x$ and $\cos x$. We know that $\sec x = \frac{1}{\cos x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$. So, $\sec^2 x = \frac{1}{\cos^2 x}$ and $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$.

Substitute these into the integral: $I = \int \frac{1}{\cos^2 x} \cdot \frac{1}{\sin^2 x} \, dx$ $I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx$

Step 2: Use the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$ in the numerator. We can replace the numerator '1' with $\sin^2 x + \cos^2 x$: $I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx$

Step 3: Split the fraction into two separate terms. $I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx$

Step 4: Simplify each term. $I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx$

Step 5: Rewrite the terms using $\sec^2 x$ and $\operatorname{cosec}^2 x$. $I = \int (\sec^2 x + \operatorname{cosec}^2 x) \, dx$

Step 6: Integrate each term separately using standard integral formulas. We know that $\int \sec^2 x \, dx = \tan x + C$ and $\int \operatorname{cosec}^2 x \, dx = -\cot x + C$. $I = \int \sec^2 x \, dx + \int \operatorname{cosec}^2 x \, dx$ $I = \tan x - \cot x + C$ where $C$ is the constant of integration.

Alternatively, the result can also be expressed as $-2 \cot(2x) + C$, as $\tan x - \cot x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} = \frac{-(\cos^2 x - \sin^2 x)}{\frac{1}{2}(2\sin x \cos x)} = \frac{-\cos(2x)}{\frac{1}{2}\sin(2x)} = -2 \cot(2x)$.