What is \(\int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x⋅\cos x}}\) dx equal to ?
Understanding the Integral Problem
The problem asks us to evaluate the indefinite integral of a trigonometric expression: \(\int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x\cdot;\cos x}}\) dx. This involves trigonometric functions and fractional exponents, requiring simplification and appropriate integration techniques.
Step-by-Step Solution to the Integral
Let the given integral be \(I\).
\[ I = \int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x\cdot\cos x}} dx \]
First, we can split the integrand into two terms:
\[ I = \int \left( \frac{(\cos x)^{1.5}}{\sqrt{\sin x\cdot\cos x}} - \frac{(\sin x)^{1.5}}{\sqrt{\sin x\cdot\cos x}} \right) dx \]
We can rewrite the terms using fractional exponents: \(\sqrt{u} = u^{0.5}\) and \(u^{1.5} = u^{1} \cdot u^{0.5}\).
The denominator is \((\sin x)^{0.5} (\cos x)^{0.5}\).
The first term becomes:
\[ \frac{(\cos x)^{1.5}}{(\sin x)^{0.5} (\cos x)^{0.5}} = \frac{(\cos x)^{1.5}}{(\cos x)^{0.5}} \cdot \frac{1}{(\sin x)^{0.5}} = (\cos x)^{1.5 - 0.5} \cdot (\sin x)^{-0.5} = (\cos x)^{1} \cdot (\sin x)^{-0.5} = \frac{\cos x}{\sqrt{\sin x}} \]
The second term becomes:
\[ \frac{(\sin x)^{1.5}}{(\sin x)^{0.5} (\cos x)^{0.5}} = \frac{(\sin x)^{1.5}}{(\sin x)^{0.5}} \cdot \frac{1}{(\cos x)^{0.5}} = (\sin x)^{1.5 - 0.5} \cdot (\cos x)^{-0.5} = (\sin x)^{1} \cdot (\cos x)^{-0.5} = \frac{\sin x}{\sqrt{\cos x}} \]
So, the integral becomes:
\[ I = \int \left( \frac{\cos x}{\sqrt{\sin x}} - \frac{\sin x}{\sqrt{\cos x}} \right) dx = \int \frac{\cos x}{\sqrt{\sin x}} dx - \int \frac{\sin x}{\sqrt{\cos x}} dx \]
Evaluating the First Integral: \(\int \frac{\cos x}{\sqrt{\sin x}} dx\)
Let \(u = \sin x\). Then, the differential \(du = \cos x\) dx.
The integral transforms to:
\[ \int \frac{1}{\sqrt{u}} du = \int u^{-0.5} du \]
Using the power rule for integration, \(\int y^n dy = \frac{y^{n+1}}{n+1} + C\) (for \(n \ne -1\)):
\[ \int u^{-0.5} du = \frac{u^{-0.5+1}}{-0.5+1} + C_1 = \frac{u^{0.5}}{0.5} + C_1 = 2\sqrt{u} + C_1 \]
Substituting back \(u = \sin x\):
\[ \int \frac{\cos x}{\sqrt{\sin x}} dx = 2\sqrt{\sin x} + C_1 \]
Evaluating the Second Integral: \(\int \frac{\sin x}{\sqrt{\cos x}} dx\)
Let \(v = \cos x\). Then, the differential \(dv = -\sin x\) dx, which means \(\sin x\) dx = -dv.
The integral transforms to:
\[ \int \frac{-dv}{\sqrt{v}} = - \int v^{-0.5} dv \]
Using the power rule for integration:
\[ - \int v^{-0.5} dv = - \left( \frac{v^{-0.5+1}}{-0.5+1} \right) + C_2 = - \left( \frac{v^{0.5}}{0.5} \right) + C_2 = -2\sqrt{v} + C_2 \]
Substituting back \(v = \cos x\):
\[ \int \frac{\sin x}{\sqrt{\cos x}} dx = -2\sqrt{\cos x} + C_2 \]
Combining the Results
Now, we combine the results of the two integrals:
\[ I = (2\sqrt{\sin x} + C_1) - (-2\sqrt{\cos x} + C_2) \]
\[ I = 2\sqrt{\sin x} + C_1 + 2\sqrt{\cos x} - C_2 \]
Combining the constants of integration \(C_1 - C_2\) into a single constant \(C\):
\[ I = 2\sqrt{\sin x} + 2\sqrt{\cos x} + C \]
Comparing with Options
Let's compare our result with the given options:
- \(\sqrt{\sin \text{x}}−\sqrt{\cos \text{x}}\) + c
- \(\sqrt{\sin \text{x}}+\sqrt{\cos \text{x}}\) + c
- \(2\sqrt{\sin \text{x}}+2 \sqrt{\cos \text{x}}\) + c
- \(\frac{1}{2}\sqrt{\sin \text{x}}+\frac{1}{2} \sqrt{\cos \text{x}}\) + c
Our calculated result is \(2\sqrt{\sin x} + 2\sqrt{\cos x} + C\), which matches Option 3.
| Step | Description | Mathematical Expression |
|---|---|---|
| 1 | Split the integral | \(\int \left( \frac{(\cos x)^{1.5}}{\sqrt{\sin x\cdot\cos x}} - \frac{(\sin x)^{1.5}}{\sqrt{\sin x\cdot\cos x}} \right) dx\) |
| 2 | Simplify the first term | \(\frac{(\cos x)^{1.5}}{(\sin x)^{0.5} (\cos x)^{0.5}} = \frac{\cos x}{\sqrt{\sin x}}\) |
| 3 | Simplify the second term | \(\frac{(\sin x)^{1.5}}{(\sin x)^{0.5} (\cos x)^{0.5}} = \frac{\sin x}{\sqrt{\cos x}}\) |
| 4 | Rewrite the integral | \(\int \frac{\cos x}{\sqrt{\sin x}} dx - \int \frac{\sin x}{\sqrt{\cos x}} dx\) |
| 5 | Integrate the first term (using \(u=\sin x\)) | \(2\sqrt{\sin x} + C_1\) |
| 6 | Integrate the second term (using \(v=\cos x\)) | \(-2\sqrt{\cos x} + C_2\) |
| 7 | Combine the results | \(2\sqrt{\sin x} + 2\sqrt{\cos x} + C\) |
Revision Table: Integral Calculus
| Concept | Description | Example/Formula |
|---|---|---|
| Indefinite Integral | The reverse process of differentiation; finds the antiderivative. | \(\int f(x) dx = F(x) + C\) (where \(F'(x) = f(x)\)) |
| Power Rule (Integration) | Rule for integrating powers of a variable. | \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) (for \(n \ne -1\)) |
| Substitution Method | A technique to simplify integrals by changing the variable. | Used when the integrand contains a function and its derivative. |
| Constant of Integration | An arbitrary constant added to indefinite integrals because the derivative of a constant is zero. | Represented by \(C\). |
Additional Information: Trigonometric Integrals
Integrating trigonometric functions often requires recognizing patterns, using identities, or applying substitution. In this problem, the presence of a function \(\sin x\) under a root in the denominator and its derivative \(\cos x\) in the numerator (and vice versa) strongly suggests using the substitution method. Rewriting fractional powers and simplifying the expression before integration is a crucial first step for many complex trigonometric integrals.
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?
Evaluation of \(\displaystyle\int{\dfrac{1-\tan x}{1+\tan x}}\ dx\) is:
- \(\int\limits_{ - 2}^2 {\left| {1 - {x^2}} \right|} dx\) is:
The value of \(\int\limits_{\pi /4}^{3\pi /4} {\frac{{dx}}{{1 + \cos x}}} \) is:
