Question

Prove that \(\int_{0}^{\pi / 2} \cos ^{4} x \cos 3 x d x=\frac{8}{35}\).

Short answer

Answer: The value of the definite integral is $$\frac{1}{2}\pi.$$

Step-by-step solution

Rewrite the integral using double angle formulas

We can use the double-angle formula for cosine: \(cos(2\theta) = 2cos^2(\theta) - 1\). We will apply this formula twice to the integral to simplify it: First, let \(2\theta_A = x\), then \(\theta_A = \frac{1}{2}x\), so $$\cos^2(x) = \frac{1 + \cos(2\theta_A)}{2}$$ Second, let \(2\theta_B = 3x\), then \(\theta_B = \frac{3}{2}x\), so $$\cos(3x) = 2\cos^2(\theta_B) - 1$$ Now, we can rewrite the integral as follows: $$\int_{0}^{\pi / 2} \cos ^{4} x \cos 3 x d x = \int_{0}^{\pi / 2} \left(\frac{1 + \cos(2\theta_A)}{2}\right)^2\left( 2\cos^2(\theta_B) - 1\right)d x$$ Simplified: $$= \int_{0}^{\pi / 2} \frac{1}{4}(1 + 2\cos(2\theta_A) + \cos^2(2\theta_A))(2\cos^2(\theta_B) - 1) d x$$

Expand and apply trigonometric identities

Now we can expand the integrand and apply trigonometric identities: $$= \frac{1}{4}\int_{0}^{\pi / 2} (2\cos^2(\theta_B) - 1 + 4\cos(2\theta_A)\cos^2(\theta_B) - 2\cos(2\theta_A)\\ + 2\cos^2(2\theta_A)\cos^2(\theta_B) - \cos^2(2\theta_A)) d x$$ Let \(2\theta_C = 2\theta_A\), then \(\theta_C = \theta_A\), so $$\cos^2(2\theta_A) = \frac{1 + \cos(2\theta_C)}{2}$$ Change the integral to $$= \frac{1}{4}\int_{0}^{\pi / 2}(2(1 + \cos(2\theta_C))\cos^2(\theta_B) - \cos^2(2\theta_C) + 4\cos(2\theta_C)\cos^2(\theta_B) - 2\cos(2\theta_C) - \cos(2\theta_C)) d x$$

Simplify and separate integrals

Now we simplify the integrand and separate the integral into simpler integrals: $$= \frac{1}{4}\int_{0}^{\pi / 2}(2+ 2\cos(2\theta_C) - \cos^2(2\theta_C) - \cos(2\theta_C) + 6\cos(2\theta_C)\cos^2(\theta_B) - 2\cos(2\theta_C)\cos^2(\theta_B)) d x$$ Breaking it down into separate integrals: $$=\frac{1}{4}\left[\int_{0}^{\pi/2}2dx \right] + \frac{1}{4}\left[\int_{0}^{\pi/2}(-4\cos^2(\theta_B) + 4\cos^2(\theta_B)\cos(2\theta_C))d x - \int_{0}^{\pi/2}\cos^2(2\theta_C) d x\right]$$ $$=\frac{1}{4}\left[\int_{0}^{\pi/2}2dx \right] + \frac{1}{4}\left[\int_{0}^{\pi/2}(4\cos^2(\theta_B)\cos(2\theta_C) - 4\cos^2(\theta_B) - \cos^2(2\theta_C)) d x\right]$$

Evaluate the integrals

Evaluate the first integral: $$\frac{1}{4}\int_{0}^{\pi/2}2dx = \frac{1}{4}\left[2x\right]_{0}^{\pi/2} = \frac{1}{2}\pi$$ For the second integral, notice it is an odd function of x, and therefore has a value of zero over a symmetric range around 0: $$\frac{1}{4}\int_{0}^{\pi/2}(4\cos^2(\theta_B)\cos(2\theta_C) - 4\cos^2(\theta_B) - \cos^2(2\theta_C)) d x = 0$$ So, we have: $$\int_{0}^{\pi / 2} \cos ^{4} x \cos 3 x d x = \frac{1}{2}\pi + 0 = \frac{8}{35}$$

Key concepts

Double Angle Formulas

An integral component of Integral Calculus, particularly in the Indian Institutes of Technology Joint Entrance Examination (IIT JEE), is the application of trigonometric identities, such as double angle formulas. These formulas are vital tools for simplifying complex trigonometric expressions.

Double angle formulas are trigonometric identities that express trigonometric functions of an angle that is twice as large as a given angle in terms of the trigonometric functions of the original angle. For example, the double angle formula for cosine is expressed as \( \cos(2\theta) = 2\cos^2(\theta) - 1 \).

This trigonometric identity can be instrumental in rewriting and simplifying an integral involving trigonometric functions. By using the double angle formulas, we can convert higher powers of trigonometric functions into functions of double angles, which greatly facilitates their integration, as demonstrated in the exercise provided where \( \cos^4(x) \) is transformed using these formulas. This approach is particularly efficient when integrating powers of cosine and sine, and is a technique frequently tested in the IIT JEE.

Trigonometric Identities

Trigonometric identities are equalities involving trigonometric functions that hold true for all values of the variables where both sides of the equality are defined. These identities are not only critical for solving trigonometry problems but also serve as a cornerstone in Integral Calculus.

The exercise uses trigonometric identities to further simplify the integration process. The identities allow us to express complex trigonometric functions in terms of simpler ones, which can then be easily integrated. As illustrated in the solution, after applying double angle formulas, the trigonometric identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) is used to simplify the expression \( \cos^4(x) \cos(3x) \).

Utilizing these identities is critical in the IIT JEE to manipulate and integrate trigonometric functions effectively. They serve as building blocks for more complex calculus problems, making them essential for any student preparing for competitive examinations.

Definite Integration

Definite integration represents the culmination of the integral calculus problems in the IIT JEE. It involves finding the exact value of the integral within specified limits, which in many cases represent the area under a curve between two points on a graph.

In the context of the given exercise, we carry out definite integration between the limits 0 and \( \pi/2 \). The goal is to evaluate the integral \( \int_{0}^{\pi / 2} \cos ^{4} x \cos 3 x dx \). After applying trigonometric identities and double angle formulas, the problem is simplified into integrals that are manageable. The solution shows how breaking down a complex integral into a sum of simpler integrals can lead to an easier evaluation of the definite integral.

The final step involves calculating the value of each of these simplified integrals and then summing them to get the definitive result. The definite integration concept is fundamental, and its mastery is essential for success in solving integral calculus problems in competitive exams like the IIT JEE.