Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{2+3 \cos y}{\sin ^{2} y} d y$$

Short answer

Question: Determine the indefinite integral of the following function, and then check your work by differentiation with respect to y: $$\int \frac{2+3 \cos y}{\sin ^{2} y} d y$$ Answer: The indefinite integral of the given function is: $$- \frac{2}{\sin y} + 3 \ln |\sin y| + C$$ And when we differentiate this with respect to y, we get back the original integrand: $$\frac{2+3\cos y}{\sin^2 y}$$

Step-by-step solution

Identifying substitution possibility

This integral looks like a case for substitution method. We can see a relationship between the numerator and the denominator, i.e., the derivative of \(\sin y\) is \(\cos y\). So, let us try substituting \(u = \sin y\).

Substituting u

Since we are using the substitution \(u = \sin y\), we will now find the derivative of \(u\) with respect to \(y\), that is: $$\frac{d u}{d y} = \cos y$$ Now, we'll solve for \(d y\): $$d y = \frac{d u}{\cos y}$$ Now we substitute \(u = \sin y\) and \(d y\) in the integral: $$\int \frac{2+3 \cos y}{u^2} \frac{d u}{\cos y} = \int \frac{2+3u}{u^2} d u$$

Splitting the integral

Now, we can split the integral into two parts and solve them separately: $$\int \frac{2 + 3u}{u^2} d u = \int \frac{2}{u^2} d u + \int \frac{3u}{u^2} d u$$

Evaluating individual integrals

Let's evaluate both integrals separately. $$\int \frac{2}{u^2} d u = 2 \int u^{-2} d u$$ $$= 2 (-u^{-1}) + C_1 = - \frac{2}{u} + C_1$$ and $$\int \frac{3u}{u^2} d u = 3 \int u^{-1} d u$$ $$= 3 (\ln |u|) + C_2 = 3 \ln |u| + C_2$$

Combining the integrals

Now, we combine both results and substitute back \(u = \sin y\): $$\int \frac{2+3 \cos y}{\sin ^{2} y} d y = - \frac{2}{\sin y} + 3 \ln |\sin y| + C$$ where \(C = C_1 + C_2\) is the constant of integration.

Checking the result

We now check our result by differentiating our integral with respect to \(y\): $$\frac{d}{d y} \left(- \frac{2}{\sin y} + 3 \ln |\sin y| + C \right)$$ Using chain rule, we get: $$\frac{d}{d y} \left(- \frac{2}{\sin y}\right) + \frac{d}{d y}\left(3 \ln |\sin y| \right) = \frac{2 \cos y}{\sin^2 y} + \frac{3\cos y}{\sin y}$$ Combining, $$= \frac{2+3\cos y}{\sin^2 y}$$ Since the derivative matches the original integrand, our integral is correct.

Key concepts

Integration by Substitution

Integration by substitution, often referred to as u-substitution, is a technique similar to the chain rule for differentiation. It's used when an integral contains a function and its derivative.

Like a puzzle, we identify a piece of the integrand that we can replace with a 'u,' making the integral easier to solve. Once we 'substitute' the hard part with something simpler, we integrate with respect to 'u' and then substitute the original expression back in at the end.

For example, in the given exercise, the substitution was made with u = sin y, simplifying the integrand significantly. When you find the appropriate 'u', the integral often becomes much more straightforward. This technique not only simplifies the integrand but sometimes it's the only clear path to the solution.

Trigonometric Integration

Trigonometric integration involves finding the integral of functions that contain trigonometric functions. These can be challenging due to the properties of sine, cosine, and other trigonometric functions.

To solve these, you might use identities to simplify them or, as in our example, integrate by substitution. The trigonometric functions often have derivatives that are also trigonometric, making them suitable for u-substitution.

Remember, mastering these integrals involves knowing your trigonometric identities well and recognizing patterns in the integrand that match these identities or their derivatives.

Chain Rule Differentiation

The chain rule is a formula for finding the derivative of a composition of functions. Think of it as finding the derivative 'link by link'.

If you have a function g inside another function f, the chain rule tells us that the derivative is found by taking the derivative of the outer function and multiplying it by the derivative of the inner function. Symbolically, this is written as \(f(g(x))' = f'(g(x)) \times g'(x)\).

When you're checking your work after integrating, as in the provided exercise, you'll often need the chain rule to differentiate correctly and verify that you've found the correct antiderivative.

Constant of Integration

The constant of integration, denoted commonly by 'C', acknowledges that there are an infinite number of antiderivatives for a given function. When you integrate, you reverse differentiation, but you can't pinpoint the exact original function without initial conditions.

This is because differentiating any constant results in zero, so when you reverse the process, you must add 'C' to account for any possible constant that was lost in differentiation. It is the way we express the general solution to an indefinite integral.

In the exercise, we saw two constants, \(C_1\) and \(C_2\), from integrating two separate functions. They combine into a single \(C\) because the actual value of the constant is unknown until we have additional conditions, such as an initial value.