Question

Evaluate the following integrals: $$ \int \frac{d x}{\left(4 x-x^{2}\right)^{3 / 2}} $$

Short answer

Based on the step-by-step solution above, evaluate the integral: $$ \int \frac{1}{(4x - x^2)^{3/2}} dx $$ Solution: $$ \sqrt{2}\sqrt{\frac{x}{4}(1 - \frac{x}{4})} + C $$

Step-by-step solution

Perform the Substitution

We substitute \(x\) with \(4\sin^2{u}\). Thus, we have: $$ x = 4\sin^2{u} $$

Find the Differential

Differentiate both sides of the equation with respect to \(u\) to find the differential: $$ \frac{dx}{du} = 8\sin{u}\cos{u} $$ Now, find \(dx\) by multiplying both sides by \(du\): $$ dx = 8\sin{u}\cos{u} du $$

Substitute \(x\) and \(dx\) into the Integral

Now, we substitute \(x = 4\sin^2{u}\) and \(dx = 8\sin{u}\cos{u} du\) into the integral: $$ \int \frac{1}{(4x - x^2)^{3/2}} dx = \int \frac{8\sin{u}\cos{u}}{(4(4\sin^2{u}) - (4\sin^2{u})^2)^{3/2}} du $$

Simplify the Integral Expression

Now, let's simplify the integral expression: $$ \int \frac{8\sin{u}\cos{u}}{(16\sin^2{u} - 16\sin^4{u})^{3/2}} du = \int \frac{8\sin{u}\cos{u}}{(16\sin^2{u}(1 - \sin^2{u}))^{3/2}} du $$ Since \(\cos^2{u} = 1 - \sin^2{u}\), the integral becomes: $$ \int \frac{8\sin{u}\cos{u}}{(16\sin^2{u}\cos^2{u})^{3/2}} du $$

Solve the Integral

Now, simplify the integral, then evaluate it: $$ \int \frac{8\sin{u}\cos{u}}{(16\sin^2{u}\cos^2{u})^{3/2}} du = \int \frac{1}{2\sqrt{2}(\sin{u}\cos{u})^{1/2}} du $$ Perform the substitution \(\sin{u}\cos{u} = v\), then we have \(\frac{1}{2} du = dv\). Thus, the integral becomes: $$ \int \frac{1}{2\sqrt{2}(v)^{1/2}} dv = \frac{1}{\sqrt{2}} \int v^{-1/2} dv $$ Now, evaluate the integral: $$ \frac{1}{\sqrt{2}} \int v^{-1/2} dv = \frac{1}{\sqrt{2}} (2v^{1/2}) + C = \sqrt{2v} + C $$ Since, \(v = \sin{u}\cos{u}\) and \(x = 4\sin^2{u}\), we have \(v = \frac{x}{4}(1 - \frac{x}{4})\). Substitute \(v\) back into the solution to get the final answer: $$ \sqrt{2}\sqrt{\frac{x}{4}(1 - \frac{x}{4})} + C $$

Key concepts

Substitution Method

The substitution method is a fundamental technique in integral calculus used to simplify complex integrals. By introducing a new variable, integrations that might seem challenging at first become more manageable.

For example, in the exercise $$\int \frac{d x}{(4 x-x^{2})^{3 / 2}}$$ the substitution $$x = 4\sin^2{u}$$ transforms the integral into a form that can be more easily solved. This change of variable also necessitates finding the differential $$dx$$ which, in this case, is obtained by differentiating the substitution equation with respect to $$u$$ and multiplying by $$du$$. The substitution method can dramatically simplify the integrand, allowing for direct integration or simpler algebraic manipulation. Remember that, after integrating with respect to the new variable, you will have to 'substitute back' to return to the original variable. This step ensures you express the solution in terms of the initial variable, as shown in the provided exercise.

Integration Techniques

Integration techniques are various strategies employed to evaluate integrals that are not immediately obvious. These methods include substitution, integration by parts, partial fraction decomposition, and trigonometric substitution, among others. Each technique has its application depending on the structure of the integrand.

In the example exercise, a combination of trigonometric substitution and algebraic simplification is used to solve a seemingly complicated integral. After substituting and simplifying, the integral becomes $$\int \frac{1}{2\sqrt{2}(\sin{u}\cos{u})^{1/2}} du$$ which involves further substitution. Mastery of these techniques is crucial for students, especially those preparing for competitive examinations like the IIT JEE, where a diverse set of problems can be presented.

Definite Integrals

Definite integrals are an extension of indefinite integrals, providing the net area under a curve between two points. Unlike indefinite integrals, which include an arbitrary constant ($$C$$), definite integrals give a precise value. While the exercise in question doesn't involve a definite integral, it's important to note the significance of bounds in evaluating definite integrals.

When dealing with definite integrals, the techniques remain similar, but you will apply the limits of integration to the new variable after substitution, if applicable. Solving definite integrals also involves understanding the properties of integrals, such as symmetry and periodicity, which are vital for solving problems efficiently. In exams like the IIT JEE, being adept at switching between indefinite and definite integrals can be particularly useful.

Integral Calculus for IIT JEE

Integral calculus is a critical topic in the IIT JEE syllabus, requiring a strong grasp of concepts and techniques to solve complex problems. Integral calculus questions in IIT JEE can range from basic anti-differentiation to applying integration to physics problems, like finding the center of mass or electric potential.

Preparation for IIT JEE demands familiarity with various integration methods, manipulative algebra, and geometrical interpretations of integrals. Regular practice with diverse problems, like the integral presented in this example, ensures that IIT JEE aspirants develop the proficiency needed to handle different problems under exam conditions. Integrals are often used in combination with other mathematical concepts, making their understanding also crucial for a comprehensive approach to solving advanced problems in the JEE.