Question

Evaluate the following integrals: (i) \(\int \operatorname{cosec}^{2} x \ln \sec x d x\). (ii) \(\int \cos x \ln (\operatorname{cosec} x+\cot x) d x\) (iii) \(\int \sin x \cdot \ln (\sec x+\tan x) d x\) (iv) \(\int \sec x \cdot \ln (\sec x+\tan x) d x\)

Short answer

Question: Evaluate the following integrals: 1. \(\int \operatorname{cosec}^2 x \ln(\sec x) dx\) 2. \(\int \cos x \ln(\operatorname{cosec} x + \cot x) dx\) 3. \(\int \sin x \ln(\sec x + \tan x) dx\) 4. \(\int \sec x \ln(\sec x + \tan x) dx\) Answer: 1. \(\int \operatorname{cosec}^2 x \ln(\sec x) dx = -\cot x (\ln\sec x) - x + C\) 2. \(\int \cos x \ln(\operatorname{cosec} x + \cot x) dx = \sin x \ln(\operatorname{cosec} x + \cot x) - \ln|\operatorname{cosec} x + \cot x| + C\) 3. \(\int \sin x \ln(\sec x + \tan x) dx = -\cos x \ln(\sec x + \tan x)+\ln|\sec x + \tan x| + C\) 4. \(\int \sec x \ln(\sec x + \tan x) dx = \sec x (\ln(\sec x + \tan x))^2 - (\sec x + \tan x)(\ln(\sec x + \tan x) - 1) + C\)

Step-by-step solution

Integral (i)

We are given \(\int \operatorname{cosec}^2 x \ln(\sec x) dx\). We can apply integration by parts to this, letting \(u = \ln(\sec x)\) and \(dv = \operatorname{cosec}^2 x\, dx\). Then, \[ du = \frac{\sec{x}\tan{x}}{\sec{x}}dx = \tan{x} dx \] and \[ v = \int \operatorname{cosec}^2 x\, dx = -\cot x + C_1 \] Now applying integration by parts, we get \[ \int \operatorname{cosec}^2 x \ln(\sec x) dx = uv - \int v\,du \] \[ = -\cot x (\ln\sec x) - \int(-\cot x)(\tan x) dx \] The remaining integral can be simplified as: \[ \int \cot x \tan x dx = \int \frac{\cos x}{\sin x}\cdot \frac{\sin x}{\cos x} dx = \int dx = x + C_2 \] Thus, the final result is: \[ \int \operatorname{cosec}^2 x \ln(\sec x) dx = -\cot x (\ln\sec x) - x + C \]

Integral (ii)

For the given integral \(\int \cos x \ln(\operatorname{cosec} x + \cot x) dx\), let the function \(u\) be \(\ln(\text{cosec}(x)+\cot(x))\) and \(dv\) be \(\cos(x) dx\). Then, we get \(du\) by taking derivative of \(u\) with respect to \(x\), which gives, \[ du = \frac{-\operatorname{cosec}(x)}{\operatorname{cosec}(x)+\cot(x)} dx. \] and \[ v = \int \cos x dx = \sin x+C_1. \] With the integration by parts formula, we have \[ \int \cos x \ln (\operatorname{cosec} x+\cot x) d x = uv - \int v\, du. \] Now, we should simplify the new integral: \[ \int \sin x \cdot \frac{-\operatorname{cosec} x}{\operatorname{cosec} x + \cot x} dx = -\int \frac{1}{\operatorname{cosec} x + \cot x} dx \] We use substitution method to simplify the above integral: Let \(y=\operatorname{cosec}x+\cot x\), then \(dy=(-\operatorname{cosec}x)\,dx\). Thus, the integral becomes, \[ -\int \frac{1}{y} dy = -\ln|y|+C_2 \] Hence, the final result of the integral is \[ \int \cos x \ln(\operatorname{cosec} x + \cot x) dx = \sin x \ln(\operatorname{cosec} x + \cot x) - \ln|\operatorname{cosec} x + \cot x| + C. \]

Integral (iii)

For the given integral \(\int \sin x \ln(\sec x + \tan x) dx\), let \(u = \ln(\sec x + \tan x)\) and \(dv = \sin x dx\). Consequently, \[ du = \frac{\sec x}{\sec x + \tan x} dx \] and \[ v = \int \sin x dx = -\cos x + C_1 \] Now, let's apply integration by parts \[ \int \sin x \ln(\sec x + \tan x) dx = uv - \int v\,du \] \[ = -(\cos x) (\ln(\sec x + \tan x)) - \int \frac{-\cos x \cdot \sec x}{\sec x + \tan x} dx \] We use the following substitution for the remaining integral, \(y = \sec x + \tan x \implies dy = (\sec x)(\sec x\tan x + 1)dx\). Thus, \[ \int \frac{-\cos x \cdot \sec x}{\sec x + \tan x} dx = -\int \frac{dy}{y}. \] Simplify the result by integrating: \[ -\int \frac{dy}{y} = -\ln|y| + C_2 =-\ln|\sec x + \tan x| + C_2. \] Thus, the final result for the given integral is \[ \int \sin x \ln(\sec x + \tan x) dx = -\cos x \ln(\sec x + \tan x)+\ln|\sec x + \tan x| + C. \]

Integral (iv)

For the given integral \(\int \sec x \ln(\sec x + \tan x) dx\), let \(u = \ln(\sec x + \tan x)\) and \(dv = \sec x dx\). Then, \[ du = \frac{\sec x}{\sec x + \tan x} dx \] and \[ v = \int \sec x dx= \ln|\sec x + \tan x| + C_1 \] Now, applying integration by parts, we get \[ \int \sec x \ln(\sec x + \tan x) dx = uv - \int v\,du = uv - \int v \cdot \frac{\sec x}{\sec x + \tan x} dx \] \[ = \sec x (\ln(\sec x + \tan x))^2 -\int \frac{\sec x\ln|\sec x + \tan x|}{\sec x + \tan x}(\sec x\tan x + \sec^2 x) dx \] Since we have \(y = \sec x + \tan x \implies dy = (\sec x)(\sec x\tan x + 1)dx \implies \sec x \cdot \frac{\sec x\tan x + 1}{y} dx = dy\), the remaining integral becomes \[ \int \frac{\sec x\ln|\sec x + \tan x|}{\sec x + \tan x}(\sec x\tan x + \sec^2 x) dx = \int \ln(y) dy \] Now, integrate and simplify: \[ \int \ln(y) dy = y(\ln y-1) + C_2 = (\sec x + \tan x)(\ln(\sec x + \tan x) - 1) + C_2 \] Thus, the final result for the given integral is \[ \int \sec x \ln(\sec x + \tan x) dx = \sec x (\ln(\sec x + \tan x))^2 - (\sec x + \tan x)(\ln(\sec x + \tan x) - 1) + C. \]