Evaluate The Integral:${ 3 \pi \int_0^5 \sin \left(\frac{t \pi}{12}\right) , Dt }$

Introduction

In this article, we will delve into the world of calculus and evaluate the given integral, which involves the sine function. The integral in question is: ${ 3 \pi \int_0^5 \sin \left(\frac{t \pi}{12}\right) , dt }$. This problem requires a thorough understanding of integration techniques, particularly the substitution method. We will break down the solution step by step, providing a clear and concise explanation of each step.

Understanding the Integral

The given integral is a definite integral, which means it has a specific upper and lower limit. In this case, the lower limit is 0 and the upper limit is 5. The integral involves the sine function, which is a fundamental trigonometric function. The sine function is periodic, meaning it repeats itself at regular intervals. In this case, the sine function is being evaluated at the argument $\frac{t \pi}{12}$.

Substitution Method

To evaluate this integral, we will use the substitution method. This method involves substituting a new variable into the integral, which simplifies the integral and makes it easier to evaluate. In this case, we will substitute $u = \frac{t \pi}{12}$. This substitution will transform the integral into a more manageable form.

Finding the Derivative of the Substitution

To find the derivative of the substitution, we will use the chain rule. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of this function is given by $f'(g(x)) \cdot g'(x)$. In this case, we have $u = \frac{t \pi}{12}$, so we need to find the derivative of $u$ with respect to $t$.

Finding the Derivative of u

To find the derivative of $u$, we will use the power rule of differentiation. The power rule states that if we have a function of the form $x^n$, then the derivative of this function is given by $n \cdot x^{n-1}$. In this case, we have $u = \frac{t \pi}{12}$, so we can rewrite this as $u = t \cdot \frac{\pi}{12}$. Using the power rule, we find that the derivative of $u$ with respect to $t$ is given by $\frac{\pi}{12}$.

Substituting the Derivative into the Integral

Now that we have found the derivative of $u$, we can substitute this into the integral. We have $du = \frac{\pi}{12} dt$, so we can rewrite the integral as: ${ 3 \pi \int_0^5 \sin(u) , du }$. This is a much simpler integral to evaluate.

Evaluating the Integral

To evaluate this integral, we will use the antiderivative of the sine function. The antiderivative of the sine function is given by $-\cos(u)$. We can use this antiderivative to evaluate the integral.

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus states that if we have a definite integral of the form $\int_a^b f(x) dx$, then the value of this integral is given by $F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. In this case, we have $\int_0^5 \sin(u) du$, so we can use the fundamental theorem of calculus to evaluate this integral.

Evaluating the Antiderivative

To evaluate the antiderivative, we will use the fact that the antiderivative of the sine function is given by $-\cos(u)$. We can substitute the upper and lower limits into the antiderivative to find the value of the integral.

Finding the Value of the Integral

To find the value of the integral, we will substitute the upper and lower limits into the antiderivative. We have $-\cos(5) - (-\cos(0))$, which simplifies to $-\cos(5) + 1$.

Simplifying the Expression

To simplify the expression, we will use the fact that the cosine function is periodic. The cosine function has a period of $2 \pi$, which means that it repeats itself every $2 \pi$ radians. In this case, we have $\cos(5)$, which is equivalent to $\cos(5 - 12)$.

Finding the Equivalent Angle

To find the equivalent angle, we will use the fact that the cosine function has a period of $2 \pi$. We can subtract multiples of $2 \pi$ from the angle to find an equivalent angle. In this case, we have $5 - 12 = -7$, which is equivalent to $-7 + 12 = 5$.

Simplifying the Expression

To simplify the expression, we will use the fact that the cosine function is an even function. This means that $\cos(-x) = \cos(x)$ for all values of $x$. In this case, we have $\cos(5)$, which is equivalent to $\cos(-5)$.

Finding the Value of the Integral

To find the value of the integral, we will substitute the simplified expression into the antiderivative. We have $-\cos(5) + 1$, which simplifies to $-(-\cos(5)) + 1$.

Simplifying the Expression

To simplify the expression, we will use the fact that the negative sign is equivalent to the positive sign. In this case, we have $-(-\cos(5))$, which simplifies to $\cos(5)$.

Finding the Value of the Integral

To find the value of the integral, we will substitute the simplified expression into the antiderivative. We have $\cos(5) + 1$, which is the final answer.

Conclusion

In this article, we evaluated the given integral using the substitution method. We found the derivative of the substitution, substituted the derivative into the integral, and evaluated the integral using the antiderivative of the sine function. We applied the fundamental theorem of calculus to find the value of the integral and simplified the expression to find the final answer. The final answer is $\boxed{\cos(5) + 1}$.

Final Answer

The final answer is $\boxed{\cos(5) + 1}$.

Introduction

In this article, we will delve into the world of calculus and evaluate the given integral, which involves the sine function. The integral in question is: ${ 3 \pi \int_0^5 \sin \left(\frac{t \pi}{12}\right) , dt }$. This problem requires a thorough understanding of integration techniques, particularly the substitution method. We will break down the solution step by step, providing a clear and concise explanation of each step.

Understanding the Integral

The given integral is a definite integral, which means it has a specific upper and lower limit. In this case, the lower limit is 0 and the upper limit is 5. The integral involves the sine function, which is a fundamental trigonometric function. The sine function is periodic, meaning it repeats itself at regular intervals. In this case, the sine function is being evaluated at the argument $\frac{t \pi}{12}$.

Substitution Method

To evaluate this integral, we will use the substitution method. This method involves substituting a new variable into the integral, which simplifies the integral and makes it easier to evaluate. In this case, we will substitute $u = \frac{t \pi}{12}$. This substitution will transform the integral into a more manageable form.

Finding the Derivative of the Substitution

To find the derivative of the substitution, we will use the chain rule. The chain rule states that if we have a composite function of the form $f(g(x))$, then the derivative of this function is given by $f'(g(x)) \cdot g'(x)$. In this case, we have $u = \frac{t \pi}{12}$, so we need to find the derivative of $u$ with respect to $t$.

Finding the Derivative of u

To find the derivative of $u$, we will use the power rule of differentiation. The power rule states that if we have a function of the form $x^n$, then the derivative of this function is given by $n \cdot x^{n-1}$. In this case, we have $u = \frac{t \pi}{12}$, so we can rewrite this as $u = t \cdot \frac{\pi}{12}$. Using the power rule, we find that the derivative of $u$ with respect to $t$ is given by $\frac{\pi}{12}$.

Substituting the Derivative into the Integral

Now that we have found the derivative of $u$, we can substitute this into the integral. We have $du = \frac{\pi}{12} dt$, so we can rewrite the integral as: ${ 3 \pi \int_0^5 \sin(u) , du }$. This is a much simpler integral to evaluate.

Evaluating the Integral

To evaluate this integral, we will use the antiderivative of the sine function. The antiderivative of the sine function is given by $-\cos(u)$. We can use this antiderivative to evaluate the integral.

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus states that if we have a definite integral of the form $\int_a^b f(x) dx$, then the value of this integral is given by $F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$. In this case, we have $\int_0^5 \sin(u) du$, so we can use the fundamental theorem of calculus to evaluate this integral.

Evaluating the Antiderivative

To evaluate the antiderivative, we will use the fact that the antiderivative of the sine function is given by $-\cos(u)$. We can substitute the upper and lower limits into the antiderivative to find the value of the integral.

Finding the Value of the Integral

To find the value of the integral, we will substitute the upper and lower limits into the antiderivative. We have $-\cos(5) - (-\cos(0))$, which simplifies to $-\cos(5) + 1$.

Simplifying the Expression

To simplify the expression, we will use the fact that the cosine function is periodic. The cosine function has a period of $2 \pi$, which means that it repeats itself every $2 \pi$ radians. In this case, we have $\cos(5)$, which is equivalent to $\cos(5 - 12)$.

Finding the Equivalent Angle

To find the equivalent angle, we will use the fact that the cosine function has a period of $2 \pi$. We can subtract multiples of $2 \pi$ from the angle to find an equivalent angle. In this case, we have $5 - 12 = -7$, which is equivalent to $-7 + 12 = 5$.

Simplifying the Expression

To simplify the expression, we will use the fact that the cosine function is an even function. This means that $\cos(-x) = \cos(x)$ for all values of $x$. In this case, we have $\cos(5)$, which is equivalent to $\cos(-5)$.

Finding the Value of the Integral

To find the value of the integral, we will substitute the simplified expression into the antiderivative. We have $-\cos(5) + 1$, which simplifies to $-(-\cos(5)) + 1$.

Simplifying the Expression

To simplify the expression, we will use the fact that the negative sign is equivalent to the positive sign. In this case, we have $-(-\cos(5))$, which simplifies to $\cos(5)$.

Finding the Value of the Integral

To find the value of the integral, we will substitute the simplified expression into the antiderivative. We have $\cos(5) + 1$, which is the final answer.

Conclusion

In this article, we evaluated the given integral using the substitution method. We found the derivative of the substitution, substituted the derivative into the integral, and evaluated the integral using the antiderivative of the sine function. We applied the fundamental theorem of calculus to find the value of the integral and simplified the expression to find the final answer. The final answer is $\boxed{\cos(5) + 1}$.

Q&A

Q: What is the given integral?

A: The given integral is ${ 3 \pi \int_0^5 \sin \left(\frac{t \pi}{12}\right) , dt }$.

Q: What is the substitution method?

A: The substitution method is a technique used to evaluate integrals by substituting a new variable into the integral, which simplifies the integral and makes it easier to evaluate.

Q: What is the derivative of the substitution?

A: The derivative of the substitution is $\frac{\pi}{12}$.

Q: What is the antiderivative of the sine function?

A: The antiderivative of the sine function is $-\cos(u)$.

Q: How do we apply the fundamental theorem of calculus?

A: We apply the fundamental theorem of calculus by substituting the upper and lower limits into the antiderivative to find the value of the integral.

Q: What is the final answer?

A: The final answer is $\boxed{\cos(5) + 1}$.

Q: What is the period of the cosine function?

A: The period of the cosine function is $2 \pi$.

Q: Is the cosine function an even function?

A: Yes, the cosine function is an even function, which means that $\cos(-x) = \cos(x)$ for all values of $x$.

Q: How do we simplify the expression?

A: We simplify the expression by using the fact that the cosine function is periodic and that the negative sign is equivalent to the positive sign.

Q: What is the equivalent angle?

A: The equivalent angle is found by subtracting multiples of $2 \pi$ from the angle.

Q: How do we find the value of the integral?

A: We find the value of the integral by substituting the upper and lower limits into the antiderivative and simplifying the expression.