Problem 13 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x\) is equal to (A) \(\sin ^{-1}(\sin x+\cos x)+c\) (B) \(\sin ^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c\) (C) \(\cos ^{-1}(\sin x+\cos x)+c\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (B) \( \sin^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c \).

Step 1: Simplify the Trigonometric Terms

Start by recognizing the expression \( \sin(2x) = 2 \sin(x) \cos(x) \), which means we have a trigonometric identity. The numerator \( \cos(x) - \sin(x) \) can be related to \( \sin(x + \frac{\pi}{4}) \), but first simplify using substitution or identities. Rearrange and simplify the expression if feasible.

Step 2: Substitution for Simplification

To tackle the integral \( \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x \), consider a substitution that simplifies \( \sin x \) and \( \cos x \). Let \( t = \sin x + \cos x \). The derivative \( dt = (\cos x - \sin x) dx \) allows us to integrate directly whenever \( \cos x - \sin x = dt \).

Step 3: Express Integral in Terms of New Variable

Rewrite the integral in terms of \( t \): \[ \int \frac{dt}{\sqrt{8 - 2t^2}}. \]Recognize this as a standard integral form for inverse trigonometric functions.

Step 4: Solve the Standard Integral

The integral \( \int \frac{dt}{\sqrt{8 - 2t^2}} \) can be solved using the substitution \( t = \sqrt{4} \sin u \). Now, using the identity \( dx = \sqrt{4}\cos u\,du \), the integral becomes \[ \int \frac{\sqrt{4} \cos u }{\sqrt{8 - 2 (4) \sin^2 u}} \, du = \int \frac{2 \cos u}{\sqrt{8 - 8 \sin^2 u}} \, du = \int du, \]which simplifies to \( u + C = \sin^{-1}\left(\frac{t}{2}\right) + C\), where \( u = \sin^{-1}\left(\frac{t}{2}\right) \).

Step 5: Back-Substitute and Simplify

Replacing \( t \) in terms of \( x \) gives us \( t = \sin x + \cos x \), so the original integral evaluates to \[ \sin^{-1}\left(\frac{\sin x + \cos x}{2}\right) + C. \] This corresponds to the closest form in the options.

Key Concepts

Inverse trigonometric functions

Inverse trigonometric functions are fascinating tools that allow us to find the angle whose trigonometric function equals a given number. For example, \( \sin^{-1}(y) \) represents the angle whose sine is \( y \), often used in solving integrals and trigonometric equations.
Whenever you see integrations involving forms like \( \int \frac{dt}{\sqrt{a^2 - x^2}} \), an inverse trigonometric function can often simplify the problem.
In our exercise, we transformed the original integral into a standard form that leads to using \( \sin^{-1} \). Recognizing these forms helps in quick problem-solving.

• Typical forms: \( \sin^{-1}(x), \cos^{-1}(x), \tan^{-1}(x) \)
• Helpful for evaluating integrals
• Critical in solving trigonometric equations

Identifying these allows bringing out the "unknown" angle and translating it back into a simpler algebraic format to solve the problem effectively.

Trigonometric identities

Trigonometric identities are like puzzle keys that can unlock and simplify complex expressions involving trigonometric functions. These identities often transform bothersome equations into something more manageable.
A key identity in the exercise is \( \sin(2x) = 2 \sin(x) \cos(x) \), which played a role in simplifying the denominator in the integral. It's essential to be familiar with:

• Pythagorean identities: \( \sin^2(x) + \cos^2(x) = 1 \)
• Double angle identities: \( \sin(2x), \cos(2x) \)
• Sum and difference formulas: \( \sin(a \pm b), \cos(a \pm b) \)

Applying these efficiently eases the path to solve integration by turning complicated expressions into reduced standard integral forms.
They aid in rearranging and sometimes directly substituting into expressions for simplification or to adjust terms for further substitution.

Substitution method

The substitution method is a powerful integration technique that involves changing the variable of integration to simplify the integral. The goal is to transform the integral into a standard form that is easier to integrate.
In our problem, we used a substitution by letting \( t = \sin x + \cos x \), which considerably simplified the integration process. The derivative \( dt = (\cos x - \sin x) dx \) allowed for a seamless substitution.
Step-by-step, here's how the substitution method helps:

• Identify a substitution that reduces complexity
• Replace variables and corresponding differentials
• Transform integral to a simpler standard form
• Integrate and then revert back to the original variable

This method becomes particularly useful when direct integration seems unfathomable due to the complex structure of the integrand. Always look for opportunities to apply a clever substitution to make the integral more approachable and to achieve a cleaner solution.