Evaluation of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\) is:
log (cos x + sin x) + c
Evaluating the Integral of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\)
The problem asks us to evaluate the indefinite integral of the function \(\dfrac{1-\tan x}{1+\tan x}\) with respect to \(x\). This type of integral often requires simplifying the integrand using trigonometric identities before performing the integration.
Simplifying the Integrand: \(\dfrac{1-\tan x}{1+\tan x}\)
We can simplify the expression \(\dfrac{1-\tan x}{1+\tan x}\) by replacing \(\tan x\) with \(\dfrac{\sin x}{\cos x}\). This is a common technique for simplifying expressions involving tangent.
Let's substitute \(\tan x = \dfrac{\sin x}{\cos x}\) into the expression:
\[ \dfrac{1-\tan x}{1+\tan x} = \dfrac{1 - \dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}} \] To simplify the complex fraction, we find a common denominator for the numerator and the denominator, which is \(\cos x\):
\[ \dfrac{1 - \dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}} = \dfrac{\dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x}}{\dfrac{\cos x}{\cos x} + \dfrac{\sin x}{\cos x}} = \dfrac{\dfrac{\cos x - \sin x}{\cos x}}{\dfrac{\cos x + \sin x}{\cos x}} \] Now, we can multiply the numerator by the reciprocal of the denominator:
\[ \dfrac{\dfrac{\cos x - \sin x}{\cos x}}{\dfrac{\cos x + \sin x}{\cos x}} = \dfrac{\cos x - \sin x}{\cos x} \times \dfrac{\cos x}{\cos x + \sin x} = \dfrac{\cos x - \sin x}{\cos x + \sin x} \] So, the integral becomes:
\[ \int{\dfrac{1-\tan x}{1+\tan x}}\ dx = \int{\dfrac{\cos x - \sin x}{\cos x + \sin x}}\ dx \]
Evaluating the Integral using Substitution
The simplified integral is \(\displaystyle\int{\dfrac{\cos x - \sin x}{\cos x + \sin x}}\ dx\). This form is suitable for a substitution method. We observe that the numerator, \(\cos x - \sin x\), is the derivative of the denominator, \(\cos x + \sin x\), up to a sign change if the numerator was \(\cos x - (-\sin x)\). Actually, the derivative of \((\cos x + \sin x)\) is \((-\sin x + \cos x)\), which is exactly the numerator.
Let's use the substitution method. Let \(u\) be the denominator:
- Let \(u = \cos x + \sin x\).
Next, we find the differential \(du\) by differentiating \(u\) with respect to \(x\):
- \[ \dfrac{du}{dx} = \dfrac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x \]
- So, \(du = (\cos x - \sin x)\ dx\).
Now, we substitute \(u\) and \(du\) into the integral:
\[ \int{\dfrac{\cos x - \sin x}{\cos x + \sin x}}\ dx \] We can see that \((\cos x - \sin x)\ dx\) is \(du\) and \((\cos x + \sin x)\) is \(u\). So the integral transforms to:
\[ \int{\dfrac{du}{u}} \] This is a standard integral form.
The integral of \(\dfrac{1}{u}\) with respect to \(u\) is \(\log |u|\). Adding the constant of integration, \(c\), we get:
\[ \int{\dfrac{du}{u}} = \log |u| + c \] Finally, we substitute back \(u = \cos x + \sin x\) to express the result in terms of \(x\):
\[ \log |u| + c = \log |\cos x + \sin x| + c \] In many cases, when options are given without absolute values, we can assume the domain where the expression inside the logarithm is positive. Thus, the result is \(\log (\cos x + \sin x) + c\).
Comparing with Options
Let's compare our result, \(\log (\cos x + \sin x) + c\), with the given options:
- log (sin x - cos x) + c
- log (sin x - cot x) + c
- log (cos x + sin x) + c
- log (cos x - cot x) + c
Our derived result matches option 3 exactly.
Revision Table: Key Concepts in Solving this Integral
| Concept | Application in this problem | Importance |
|---|---|---|
| Trigonometric Identity \(\tan x = \frac{\sin x}{\cos x}\) | Used to rewrite the integrand in terms of sine and cosine for simplification. | Essential first step to make the integral manageable. |
| Simplifying Complex Fractions | Used to transform \(\dfrac{1 - \tan x}{1 + \tan x}\) into \(\dfrac{\cos x - \sin x}{\cos x + \sin x}\). | Crucial algebraic step after applying the trigonometric identity. |
| Substitution Method (u-substitution) | Used when the integrand is in the form \(\dfrac{f'(x)}{f(x)}\) or can be made into that form. Here, \(f(x) = \cos x + \sin x\) and \(f'(x) = \cos x - \sin x\). | A fundamental technique for integrating functions where a part of the integrand is the derivative of another part. |
| Standard Integral \(\int \frac{1}{u} du\) | The substitution transforms the complex integral into this basic form. | Knowledge of basic integral formulas is necessary for the final step. |
Additional Information: The Form \(\int \frac{f'(x)}{f(x)} dx\)
The integral we solved is a classic example of the form \(\int \frac{f'(x)}{f(x)} dx\). When an integral is in this form, where the numerator is the derivative of the denominator, the integration result is directly given by the natural logarithm of the absolute value of the denominator plus the constant of integration.
Mathematically:
\[ \int \dfrac{f'(x)}{f(x)} dx = \log |f(x)| + C \]
In our problem, we identified that if \(f(x) = \cos x + \sin x\), then \(f'(x) = \dfrac{d}{dx}(\cos x + \sin x) = -\sin x + \cos x = \cos x - \sin x\). The integral was \(\displaystyle\int{\dfrac{\cos x - \sin x}{\cos x + \sin x}}\ dx\), which perfectly fits the form \(\int \dfrac{f'(x)}{f(x)} dx\). Therefore, the result is \(\log |\cos x + \sin x| + C\), confirming our step-by-step calculation.
Understanding this standard form can help quickly identify and solve similar integrals without explicitly performing the u-substitution steps every time, though performing the steps as shown in the solution is crucial for understanding the process and for cases that aren't exact matches to standard forms.
Important Questions from Indefinite Integrals
The value of \(\rm \int \frac{(x+1)}{x(xe^x + 1)}\ dx\) is equal to:
What is ∫ (e log x + sin x) cos x dx equal to?
- \(\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|} dx\) is:
The value of \(\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \) is:
The value of \(\int_0^\pi {{{\sin }^6}} x \cdot {\cos ^5}x dx\) is:
