Integrals of Basic Trig Functions: Videos & Practice Problems

In the study of integrals, particularly with trigonometric functions, it's essential to understand that the process of integration is essentially the reverse of differentiation. This principle allows us to derive integrals of trigonometric functions by recalling their derivatives. For instance, the integral of the cosine function, \(\int \cos(x) \, dx\), is the sine function, expressed as \(\sin(x) + C\), where C is the constant of integration. Similarly, the integral of the sine function, \(\int \sin(x) \, dx\), results in - \cos(x) + C\).

When integrating expressions that involve multiple trigonometric functions, we can apply the sum and difference rule. For example, consider the integral \(\int (3 \sin(x) + 2 \cos(x)) \, dx\). This can be separated into two integrals: 3 \int \sin(x) \, dx + 2 \int \cos(x) \, dx\. By applying the constant multiple rule, we can factor out the constants, leading to -3 \cos(x) + 2 \sin(x) + C\ as the final result.

In more complex cases, such as \(\int (7 \sec(x) \tan(x) - \csc^2(x)) \, dx\), we again utilize the sum and difference rule to split the integral. This gives us 7 \int \sec(x) \tan(x) \, dx - \int \csc^2(x) \, dx\. Recognizing that the derivative of \(\sec(x)\) is \(\sec(x) \tan(x)\), we find that \(\int \sec(x) \tan(x) \, dx = \sec(x)\). For the integral of \(\csc^2(x)\), knowing that the derivative of \(\cot(x)\) is - \csc^2(x)\ allows us to conclude that \(\int \csc^2(x) \, dx = -\cot(x)\). Thus, the complete solution simplifies to 7 \sec(x) + \cot(x) + C\.

To effectively tackle integrals involving trigonometric functions, it is crucial to memorize the derivatives of common trigonometric functions. This knowledge enables the application of reverse logic to derive their integrals, making the process more intuitive and manageable. For example, the integral of \(\sec^2(x)\) yields \(\tan(x) + C\), while the integral of \(-\csc(x) \cot(x)\) results in \(\csc(x) + C\). By mastering these relationships, students can confidently approach a variety of integral problems involving trigonometric functions.

To verify an indefinite integral, we essentially reverse the process of differentiation. Given an integrand, we can check the correctness of its antiderivative by differentiating the proposed solution and confirming that it yields the original function.

Consider the indefinite integral of a function, where the proposed solution is expressed as:

$$ F(x) = 2 \sin(\sqrt{x}) + C $$

To verify this, we differentiate \( F(x) \). The derivative of a constant \( C \) is zero, so we focus on differentiating \( 2 \sin(\sqrt{x}) \).

Using the chain rule, we find the derivative of \( \sin(u) \) where \( u = \sqrt{x} \). The derivative of \( \sin(u) \) is \( \cos(u) \), thus:

$$ \frac{d}{dx}[2 \sin(\sqrt{x})] = 2 \cos(\sqrt{x}) \cdot \frac{d}{dx}[\sqrt{x}] $$

Next, we compute the derivative of \( \sqrt{x} \). Recognizing that \( \sqrt{x} = x^{1/2} \), we apply the power rule:

$$ \frac{d}{dx}[x^{1/2}] = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} $$

Substituting this back into our derivative calculation gives:

$$ \frac{d}{dx}[2 \sin(\sqrt{x})] = 2 \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} $$

Upon simplifying, the factor of 2 cancels out, resulting in:

$$ \cos(\sqrt{x}) \cdot \frac{1}{\sqrt{x}} $$

This expression matches the original integrand, confirming that the differentiation of the proposed solution yields the integrand. Therefore, we have verified that:

$$ \int \cos(\sqrt{x}) \, dx = 2 \sin(\sqrt{x}) + C $$

In summary, verifying an indefinite integral involves differentiating the proposed antiderivative and ensuring it equals the original integrand, illustrating the fundamental relationship between integration and differentiation.

To evaluate the indefinite integral of the product of cosine and sine raised to the ninety-ninth power, we can utilize substitution effectively. The integral we are considering is:

∫ cos(t) sin⁹⁹(t) dt.

First, we choose a substitution. A good candidate for u is the sine function, since it is raised to a power. Let:

u = sin(t).

Then, we find the differential du. The derivative of sin(t) is cos(t), so:

du = cos(t) dt.

Now, we can rewrite the integral in terms of u and du. Rearranging the original integral gives us:

∫ sin⁹⁹(t) cos(t) dt = ∫ u⁹⁹ du.

This integral can now be evaluated using the power rule. According to the power rule, when integrating uⁿ, we add one to the exponent and divide by the new exponent:

∫ u⁹⁹ du = (1/100) u¹⁰⁰ + C.

Substituting back for u, we have:

(1/100) sin¹⁰⁰(t) + C.

To present this in a more conventional format, we can write the final answer as:

(1/100) sin¹⁰⁰(t) + C.

This result represents the indefinite integral of the original function, showcasing the power of substitution in simplifying the integration process.