Problem 20 - Calculus 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

sIn first integral on R.H.S. put \(\tan x=t\) and in sceond integral put \(\sqrt{2} \sin x=u\) 1.e., \(\sec ^{2} d x=d t\) and \(\cos x d x=\frac{d u}{\sqrt{2}}\) \(\therefore I=\int \frac{d t}{\sqrt{\left(t^{2}-1\right)}}-\frac{2}{\sqrt{2}} \int \frac{d u}{\sqrt{\left(u^{2}-1\right)}}\) \(=\ln \left|t+\sqrt{\left(t^{2}-1\right)}\right|-\sqrt{2} \ln \left|u+\sqrt{\left(u^{2}-1\right)}\right|+c\) \(=\ln \left|\tan x+\sqrt{\left(\tan ^{2} x-1\right)}\right|-\sqrt{2} \ln \mid \sqrt{2} \sin\) \(x+\sqrt{\left(2 \sin ^{2} x-1\right)} \mid+c\) (ii) Evaluate \(\int \sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)} d x\) Solution Lct \(I=\int \sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)} d x=\int \frac{\left(\operatorname{cosec}^{2} x \pm a\right)}{\sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)}} d x\) \(=\int \frac{\operatorname{cosec}^{2} x d x}{\sqrt{\left(\operatorname{coscc}^{2} x \pm a\right)}} \pm a \int \frac{d x}{\sqrt{\left(\operatorname{cosec}^{3} x \pm a\right)}}\) \(=\int \frac{\operatorname{coscc}^{2} x d x}{\sqrt{(1 \pm a)}+\cot ^{2} x} \pm a \int \frac{\sin x d x}{\sqrt{1 \pm a\left(1-\cos ^{2} x\right)}}\) In first integral on R.H.S. put \(\cot x=t\) and in second integral put \(\cos x=u\). (iii) Evaluate \(\int \sqrt{\left(\operatorname{coscc}^{2} x+3\right)} d x\)omc Important Questions with Solution: (i) Evaluate \(\int \sqrt{\left(\operatorname{scc}^{2} x-2\right)} d x\) Solution $$ \begin{aligned} \text { Let } I &=\int \sqrt{\left(\sec ^{2} x-2\right)} d x=\int \frac{\left(\sec ^{2} x-2\right)}{\sqrt{\left(\sec ^{2} x-2\right)}} d x \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\sec ^{2} x-2\right)}}-2 \int \frac{d x}{\sqrt{\left(\sec ^{2} x-2\right)}} \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\tan ^{2} x-1\right)}}-2 \int \frac{\cos x d x}{\sqrt{1-2\left(1-\sin ^{2} x\right)}} \\ &=\int \frac{\sec ^{2} x d x}{\sqrt{\left(\tan ^{2} x-1\right)}}-2 \int \frac{\cos x d x}{\sqrt{(\sqrt{2} \sin x)^{2}-1}} \end{aligned} $$

Step-by-Step Solution

Answer

Use substitutions \( \tan x = t \) and \( \sqrt{2} \sin x = u \), then apply antiderivatives of \( \frac{1}{\sqrt{t^2-1}} \) and \( \frac{1}{\sqrt{u^2-1}} \).

Step 1: Identify Substitution

The question provides two separate integrals where substitution methods are necessary. For the first integral, use the substitution \( \tan x = t \) with \( \sec^2 x \, dx = dt \). For the second integral, use \( \sqrt{2} \sin x = u \) with \( \cos x \, dx = \frac{du}{\sqrt{2}} \).

Step 2: Transform the Integrals

For the first integral, \( \int \sec^2 x \, dx = \int dt \), which converts to \( \int \frac{dt}{\sqrt{t^2-1}} \). For the second integral, replace using the substitution to get \( -\frac{2}{\sqrt{2}} \int \frac{du}{\sqrt{u^2-1}} \).

Step 3: Find the Antiderivatives

The antiderivative of \( \int \frac{dt}{\sqrt{t^2 - 1}} \) is \( \ln \left| t + \sqrt{t^2 - 1} \right| + C_1 \). For \( \int \frac{du}{\sqrt{u^2 - 1}} \), it is \( \ln \left| u + \sqrt{u^2 - 1} \right| + C_2 \).

Step 4: Integrate Each Term

Apply the respective antiderivatives to each integral in the expression: \( \ln \left| \tan x + \sqrt{\tan^2 x - 1} \right| - \sqrt{2} \ln \left| \sqrt{2} \sin x + \sqrt{2 \sin^2 x - 1} \right| + c \), where \( c \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is an important technique to simplify complex integrals into more manageable forms. This method involves replacing part of the integral with a new variable, which can make the process of integration easier. In our exercise, we encounter two primary substitutions:

• First, substitute \( \tan x = t \), leading to the differential \( \sec^2 x \, dx = dt \). This substitution is useful because it leverages the derivative relationship where the integral involving \( \sec^2 x \) simplifies significantly.
• Secondly, substitute \( \sqrt{2} \sin x = u \), giving us \( \cos x \, dx = \frac{du}{\sqrt{2}} \). This manipulates the integral involving \( \cos x \) by utilizing the derivative properties of trigonometric functions, offering a new perspective on simplifying the integral.

The essence is to reduce the complexity by identifying the integral's components that can be swapped with simpler forms, transforming the integration process into evaluating more familiar functions.

Trigonometric Integrals

Trigonometric integrals often involve functions like sine, cosine, tangent, and their reciprocals. These integrals can initially seem daunting due to the oscillatory nature of trigonometric functions. However, identifying patterns and utilizing substitution methods or other identities can simplify these exercises significantly.

In our example, both \( \tan x \) and \( \sin x \) are key trigonometric functions that transform the integrals into simpler radical expressions. The integral involving \( \tan x \) converts into \( \int \frac{dt}{\sqrt{t^2-1}} \), a form that can be easier to handle due to its common structure seen in calculus.

By recognizing how trigonometric identities and substitutions can reformulate these problems, you focus on evaluating more straightforward integrals, leading directly to the solution. Understanding the connections between trigonometric functions through these techniques offers a streamlined path through potentially complex calculus challenges.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions which represent the inverse operation of differentiation. In this particular exercise, we derive antiderivatives from expressions like \( \int \frac{dt}{\sqrt{t^2 - 1}} \) and \( \int \frac{du}{\sqrt{u^2 - 1}} \).

Specific antiderivatives that are crucial here include:

• For \( \int \frac{dt}{\sqrt{t^2 - 1}} \), the result is \( \ln \left| t + \sqrt{t^2 - 1} \right| + C \). This antiderivative represents the hyperbolic nature of the integral's structure.
• Similarly, \( \int \frac{du}{\sqrt{u^2 - 1}} \) results in the antiderivative \( \ln \left| u + \sqrt{u^2 - 1} \right| + C \), highlighting a similar transformation pattern.

Each antiderivative corresponds to a function whose derivative will return to the original integrand. The integration process often results in constants of integration, represented as 'C', which are vital in determining the family of possible solutions. Understanding these antiderivatives allows us to solve integrals more efficiently and accurately, offering insight into the relationships being evaluated.