Integrals of Basic Trig Functions: Videos & Practice Problems

Integrating trigonometric functions involves using the reverse of their derivatives. For example, the integral of $\cos \left(x\right)$ is $\sin \left(x\right)$ + c, while the integral of $\sin \left(x\right)$ is -$\cos \left(x\right)$ + c. For more complex functions, apply the sum and difference rule and the constant multiple rule. Remembering these relationships aids in solving integrals involving secant, tangent, and cosecant functions effectively.

Integrals Resulting in Basic Trig Functions

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.

Find $g\left(\theta\right)$ by evaluating the following indefinite integral.
$g\left(\theta\right)=\int^{}\left(5\sec^2\theta-2\csc^2\theta\right)d\theta$

• A. $g\left(\theta\right)=5\tan\theta+2\cot\theta+C$
• B. $g\left(\theta\right)=5\tan\theta-2\cot\theta+C$
• C. $g\left(\theta\right)=5\cot\theta+2\tan\theta+C$
• D. $g\left(\theta\right)=5\cot\theta-2\tan\theta+C$

Answer: (A)

Find $g\left(x\right)$ by evaluating the following indefinite integral.
$g\left(x\right)=\int\left(\sin^2x-100\csc x\cot x+\cos^2x\right)dx$

• A. $g\left(x\right)=100\csc x+x+C$
• B. $g\left(x\right)=-100\csc x+x+C$
• C. $g\left(x\right)=100\csc x+C$
• D. $g\left(x\right)=-100\csc x+C$

Answer: (A)

Integrals Resulting in Basic Trig Functions Example 1

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.

Example 2: Substitution Involving Trig Functions

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.

Evaluate the indefinite integral.
$\int_{}^{}\theta\bullet sec^2\left(5\theta^2+1\right)d\theta$

• A. $10\tan\left(5\theta^2+1\right)+C$
• B. $\frac{1}{10}\tan\left(5\theta^2+1\right)+C$
• C.
  $10sec^2\left(5\theta+1\right)+C$
• D. $\frac{1}{20}\theta^2\tan\left(5\theta^2+1\right)+C$

Answer: (B)

Evaluate the definite integral.
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan y$ $sec^2ydy$

• A. 0.308
• B. $\pi$
• C. 0
• D. $\frac{\pi}{4}$

Answer: (C)

Here’s what students ask on this topic:

What is the integral of sin(x) and cos(x)?

The integral of $\sin \left(x\right)$ is $-\cos \left(x\right)+c$, where $c$ is the constant of integration. Similarly, the integral of $\cos \left(x\right)$ is $\sin \left(x\right)+c$. These results are derived by reversing the derivative process. For example, since the derivative of $\cos \left(x\right)$ is $-\sin \left(x\right)$, the integral of $\sin \left(x\right)$ must be $-\cos \left(x\right)$.

How do you integrate sec(x)tan(x) and cosec(x)cot(x)?

The integral of $\sec \left(x\right)\tan \left(x\right)$ is $\sec \left(x\right)+c$, and the integral of $\csc \left(x\right)\cot \left(x\right)$ is $-\csc \left(x\right)+c$. These results are based on the reverse derivative process. For instance, since the derivative of $\sec \left(x\right)$ is $\sec \left(x\right)\tan \left(x\right)$, the integral of $\sec \left(x\right)\tan \left(x\right)$ must be $\sec \left(x\right)$.

What is the integral of sec²(x) and csc²(x)?

The integral of $\sec ²\left(x\right)$ is $\tan \left(x\right)+c$, and the integral of $\csc ²\left(x\right)$ is $-\cot \left(x\right)+c$. These results are derived from the reverse derivative process. For example, since the derivative of $\tan \left(x\right)$ is $\sec ²\left(x\right)$, the integral of $\sec ²\left(x\right)$ must be $\tan \left(x\right)$.

How do you apply the sum and difference rule to integrate trigonometric functions?

The sum and difference rule allows you to split an integral into separate terms. For example, to integrate $3\sin \left(x\right)+2\cos \left(x\right)$, you can write it as $3\int \sin \left(x\right)dx+2\int \cos \left(x\right)dx$. Then, apply the constant multiple rule to pull out constants, resulting in $3(-\cos (x\left)\right)+2\sin \left(x\right)+c$. This simplifies to $-3\cos \left(x\right)+2\sin \left(x\right)+c$.

Why is it important to memorize common trigonometric integrals?

Memorizing common trigonometric integrals, such as $\int \sin \left(x\right)dx=-\cos \left(x\right)+c$ and $\int \cos \left(x\right)dx=\sin \left(x\right)+c$, is crucial because it saves time during problem-solving and helps in tackling more complex integrals. These integrals often appear in calculus problems, and knowing them by heart allows you to focus on applying rules like the sum and difference rule or substitution, rather than deriving basic results repeatedly. Additionally, understanding these integrals reinforces your grasp of the relationship between derivatives and integrals, which is fundamental in calculus.