Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{d x}{\cos x+\sqrt{3} \sin x}\) equals (A) \(\frac{1}{2} \operatorname{logtan}\left(\frac{x}{2}+\frac{\pi}{12}\right)+\mathrm{c}\) (B) \(\frac{1}{2} \log \tan \left(\frac{x}{2}-\frac{\pi}{12}\right)+\mathrm{c}\) (C) \(\log \tan \left(\frac{-}{2}-\frac{\underline{\phantom{xx}}}{12}\right)+\mathrm{c}\) (D) \(\log \tan \left(\frac{x}{2}-\frac{\pi}{12}\right)+\mathrm{c}\)

Step-by-Step Solution

Answer

Option (B) is correct.

Step 1: Recognize the Trigonometric Form

Observe the integral \(\int \frac{dx}{\cos x + \sqrt{3} \sin x}\). The denominator is in the form \(\cos x + a \sin x\), where \(a = \sqrt{3}\).

Step 2: Simplify the Denominator

Rewrite \(\cos x + \sqrt{3} \sin x\) using the trigonometric identity:\[\cos x + \sqrt{3} \sin x = 2 \left(\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x\right)\]This is equivalent to:\[2 \left(\cos \frac{\pi}{3} \cos x + \sin \frac{\pi}{3} \sin x\right)\]Therefore, the expression simplifies to \(2 \cos\left(x - \frac{\pi}{3}\right)\).

Step 3: Substitute and Simplify the Integral

Now substitute the simplified expression back into the integral:\[\int \frac{dx}{\cos x + \sqrt{3} \sin x} = \int \frac{dx}{2 \cos\left(x - \frac{\pi}{3}\right)}\]\[= \frac{1}{2} \int \sec\left(x - \frac{\pi}{3}\right) dx\]

Step 4: Integrate Using Formula

The integral \(\int \sec u \, du = \log |\tan \frac{u}{2} + \frac{1}{2}| + C\). Apply this to our expression:\[\frac{1}{2} \int \sec\left(x - \frac{\pi}{3}\right) dx = \frac{1}{2} \log \left| \tan \left(\frac{x}{2} - \frac{\pi}{12}\right) \right| + C\]

Step 5: Select the Correct Option

Compare the derived integral expression with the given options, and recognize it matches option (B):\(\frac{1}{2} \log \tan \left(\frac{x}{2} - \frac{\pi}{12}\right) + C\).

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve the integration of functions containing trigonometric expressions. A fundamental aspect of solving such integrals is recognizing when a given trigonometric identity can be used to simplify the expression. In the exercise provided, the integrand is \( \frac{1}{\cos x + \sqrt{3} \sin x} \). By identifying this denominator as part of a known trigonometric identity, we can simplify it.To handle \( \cos x + a \sin x \, (where \, a = \sqrt{3}) \, \), apply the identity \[\cos x + a \sin x = R \cos(x - \alpha)\]where \[R = \sqrt{1 + a^2}\]and\[\tan \alpha = a\].In this problem:

• Perform the calculations:\[\cos x + \sqrt{3} \sin x = 2 \cos \left(x - \frac{\pi}{3}\right)\].
• This simplifies the denominator and makes it easier to approach the integration.

Understanding these identities and transformations is key to integrating trigonometric functions.

Integration Techniques

Integration techniques are strategies used to compute integrals, especially when they are not straightforward. In this exercise, simplifying the denominator in the integral is a critical step. Once the integrand is simplified, advanced techniques can be applied effectively.In our exercise:

• Initially, we identified and simplified the trigonometric form.
• Rewriting \[\frac{1}{\cos x + \sqrt{3} \sin x}\]as \[\frac{1}{2 \cos(x - \frac{\pi}{3})}\]helps to apply a known integral formula.
• Using the formula \(\int \sec u \, du = \log |\tan \frac{u}{2} \pm \frac{1}{2}| + C\) allows us to integrate effectively.

These techniques demonstrate how transforming and recognizing integrals can simplify the solution.

Definite Integrals

While the given problem focuses on an indefinite integral, definite integrals are another critical component of integral calculus. They extend the concept by finding the area under a curve from one point to another. Unlike indefinite integrals, which have a constant of integration \(C\), definite integrals result in a specific numerical value. The notational difference is clear as definite integrals use limits of integration specified at the top and bottom of the integral sign. Here are the steps:

• Evaluate the antiderivative using similar techniques as those in indefinite integration.
• Apply the Fundamental Theorem of Calculus. It states that if \(F\) is an antiderivative of \(f\), then\[\int_{a}^{b} f(x) \, dx = F(b) - F(a).\]

Incorporating the knowledge of transforming indefinite to definite integrals broadens the practical applications of calculus in real-world problems such as physics and engineering.