Question

Evaluate the following integrals : (i) \(\int \frac{1}{x \sqrt{\left(1+x^{6}\right)}} d x\) (ii) \(\int \frac{x d x}{(p x+q)^{3 / 2}}\) (iii) \(\int \frac{x^{5}}{\sqrt{\left(1+x^{2}\right.}} d x\) (iv) \(\int \frac{d x}{x^{3} \sqrt{x^{2}+1}}\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{1}{x \sqrt{\left(1+x^{6}\right)}} dx\) (ii) \(\int \frac{x dx}{(px+q)^{3 / 2}}\) (iii) \(\int \frac{x^5}{\sqrt{\left(1+x^{2}\right)}} dx\) (iv) \(\int \frac{dx}{x^3 \sqrt{x^2+1}}\) Answer: (i) \(\frac{1}{2} (\ln|x^2 + \sqrt{x^4+x^6}| + C)\) (ii) \(-\frac{2}{p}(px + q)^{-1/2} + C\) (iii) \(\frac{1}{2} (2x^{5}\sqrt{1+x^2} - \frac{10}{3}x^3\sqrt{1+x^2} + C)\) (iv) \(-\frac{1}{2} \frac{1}{x^3\sqrt{x^2+1}} + C\)

Step-by-step solution

Substitution

Let \(u = x^6\). Then, \(d u = 6 x^{5} dx\). We also need to find the new limits of integration in terms of u. Notice that we only need to make the change for x term: \(x=\sqrt[6]{u}\) Now, we can rewrite the integral: $$\int \frac{1}{x \sqrt{\left(1+x^{6}\right)}} dx = \frac{1}{6} \int \frac{1}{\sqrt[6]{u} \sqrt{1+u}} du$$

Integration

Now, evaluate the integral: $$\frac{1}{6} \int \frac{1}{\sqrt[6]{u} \sqrt{1+u}} du = \frac{1}{6} \int \frac{u^{-1/6}}{\sqrt{u^{1/3}+u^{2/3}}} du$$ Let \(v = u^{1/3}\), then, \(dv = \frac{1}{3}u^{-2/3}du\). Thus, the integral becomes: $$\frac{1}{2} \int \frac{v^{-1} dv}{\sqrt{1+v}}$$ Now, we can integrate using the substitution formula: $$\frac{1}{2} \int \frac{v^{-1} dv}{\sqrt{1+v}} = \frac{1}{2} \int \frac{dv}{v \sqrt{v+1}}$$

Final Integral and Solution

Now, integrate and reverse the substitutions: $$\frac{1}{2} \int \frac{dv}{v \sqrt{v+1}} = \frac{1}{2} (\ln|v + \sqrt{v+1}| + C)$$ Replacing \(v\) with \(u^{1/3}\): $$\frac{1}{2} (\ln|u^{1/3} + \sqrt{u^{2/3}+u}| + C)$$ Replacing \(u\) with \(x^6\): $$\frac{1}{2} (\ln|x^2 + \sqrt{x^4+x^6}| + C)$$ ## Evaluate Integral (ii):##

Substitution

Let \(t = px + q\). Then, \(dt = pdx\). Now, we can rewrite the integral: $$\int \frac{x dx}{(px+q)^{3 / 2}} = \frac{1}{p} \int \frac{1}{t^{3 / 2}} dt$$

Integration

Now, integrate by finding the antiderivative: $$\frac{1}{p} \int \frac{1}{t^{3 / 2}} dt = \frac{1}{p} \cdot -2t^{-1/2} + C = -\frac{2}{p}t^{-1/2} + C$$

Final Solution

Reverse the substitution by replacing \(t\) with \(px + q\): $$-\frac{2}{p}(px + q)^{-1/2} + C$$ ## Evaluate Integral (iii):##

Substitution

Let \(w = x^2\), then \(dw = 2xdx\). Rewrite the integral: $$\int \frac{x^5}{\sqrt{\left(1+x^{2}\right)}} dx = \frac{1}{2} \int \frac{w^{2}}{\sqrt{1+w}} dw$$

Integration

Integrating by parts with \(u = w^2\) and \(dv = w^{-1/2} dw\) gives: $$\frac{1}{2} \int \frac{w^{2}}{\sqrt{1+w}} dw = \frac{1}{2} (2w^{5/2} - \frac{10}{3}w^{3/2} + C)$$

Final Solution

Reverse the substitution by replacing \(w\) with \(x^2\): $$\frac{1}{2} (2x^{5}\sqrt{1+x^2} - \frac{10}{3}x^3\sqrt{1+x^2} + C)$$ ## Evaluate Integral (iv):##

Substitution

Let \(z = x^2\), then \(dz = 2xdx\). Rewrite the integral: $$\int \frac{dx}{x^3 \sqrt{x^2 + 1}} = \frac{1}{2} \int \frac{dz}{(z^2)\sqrt{z+1}}$$

Integration

Now, integrate by finding the antiderivative: $$\frac{1}{2} \int \frac{dz}{(z^2) \sqrt{z + 1}} = -\frac{1}{2} \frac{1}{z\sqrt{z+1}} + C$$

Final Solution

Reverse the substitution by replacing \(z\) with \(x^2\): $$-\frac{1}{2} \frac{1}{x^3\sqrt{x^2+1}} + C$$

Key concepts

Integral Calculus for IIT JEE

When preparing for one of India's most challenging engineering entrance exams, the IIT JEE, a strong command of integral calculus is a non-negotiable. It involves sophisticated problem-solving skills that encapsulate a variety of integration methods. Integral calculus problems often involve functions and shapes that can be both intricate and diverse.

Thus, the syllabus requires students to master techniques like integration by substitution, integration by parts, partial fractions, and reasonable approximation methods for complex integrals. The questions typically challenge the student's ability to transform the given integrals into simpler forms or fit them into recognizable patterns for easier evaluation.

Moreover, conceptual understanding is paramount; students need to grasp the fundamental theorems of calculus, the significance of definite and indefinite integrals, and their applications in areas like physics, engineering, and statistics. Exercises such as determining the areas under curves or the volumes of solids by integration are standard questions in the IIT JEE papers.

Therefore, a methodical approach, combined with regular practice of varied problem sets, like those given in textbook exercises, can build the proficiency necessary to tackle IIT JEE calculus problems with confidence.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a method analogous to the chain rule in differentiation. It's a powerful tool for solving integrals, especially when faced with a complicated function whose antiderivative is not readily apparent.

The strategy involves choosing a substitution that simplifies the integral into a more manageable form. This might mean transforming a trigonometric integral into an algebraic one, or turning a complex rational expression into a simpler polynomial.

Steps of Integration by Substitution:

• Pick a part of the integral to substitute with a new variable, ideally making the integral simpler.
• Differentiate the chosen part to find the differential change.
• Rewrite the integral in terms of the new variable and its differential.
• Perform the integration on the transformed integral.
• Substitute back in the original variable if necessary, as in the case of indefinite integrals.

The original exercise demonstrates this technique beautifully, going from a challenging problem to a simpler form through several layers of substitution, thus breaking down intimidating integrals into step-by-step manageable solutions.

Integration Techniques

Integration techniques are the different methods used to find antiderivatives or evaluate integrals, and they are crucial in solving diverse calculus problems. Each technique has unique applications and is chosen based on the function we're integrating.

Some of the commonly used techniques include:

• Integration by Parts: This technique utilizes the product rule of differentiation in reverse and is ideal for integrals that are products of two functions.
• Trigonometric Substitution: For integrals involving sqrt(a^2 - x^2), sqrt(a^2 + x^2), or sqrt(x^2 - a^2), trigonometric identities can simplify the functions into a form where regular integration methods apply.
• Partial Fraction Decomposition: This approach is useful for rational functions, where the integral can be broken down into a sum of simpler fractions.
• Numerical Integration: When an integral cannot be solved analytically, numerical techniques like the Trapezoidal Rule or Simpson's Rule may come in handy.

Integral calculus questions often require a blend of these techniques, as in the step-by-step solutions provided. Mastery of these methods can significantly amplify a student’s problem-solving arsenal.

Indefinite Integrals

Indefinite integrals, often referred to as antiderivatives, represent a class of integrals that do not have specified limits of integration. The general form of an indefinite integral of a function f(x) is written as \(\int f(x) dx\) and is interpreted as the family of all functions whose derivative is f(x).

Some characteristics of indefinite integrals include:

• The presence of an arbitrary constant of integration, denoted as C, which accounts for all possible antiderivatives.
• They are useful in problems regarding motion, where the goal is to find an expression for velocity or position from a known acceleration function.
• Indefinite integrals do not have a definite numerical value until the context of the problem provides initial conditions or boundaries.

Moreover, solving indefinite integrals often involves techniques such as substitution or integration by parts, as exemplified in the step-by-step solutions. Understanding indefinite integrals is fundamental in integral calculus, and students need to practice these along with their definite counterparts to develop a well-rounded understanding of the subject.