∫0π4sin(4t)dt&nbsp;

To solve the integral ∫0π4sin(4t)dt, we can use the basic properties of integrals and the trigonometric identity sin(2θ)=2sin(θ)cos(θ).Let's begin by applying the substitution u=4t. This implies du=4dt, or equivalently, dt=14du. We also need to determine the new limits of integration when we substitute t with u. When t=0, we have u=4·0=0, and when t=π4, we have u=4·π4=π. Therefore, the integral becomes:∫0π4sin(4t)dt=14∫0πsin(u)duNext, we can evaluate the integral of sin(u) with respect to u. Using the antiderivative of sin(u), which is −cos(u), we get:14∫0πsin(u)du=14[−cos(u)]0πNow, we substitute the limits of integration:14[−cos(u)]0π=14[−cos(π)−(−cos(0))]Since cos(π)=−1 and cos(0)=1, we simplify further:14[−cos(π)−(−cos(0))]=14[−(−1)−1]Now, we can simplify the expression inside the brackets:14[−(−1)−1]=14[1−1]=14·0=0Therefore, the solution to the integral ∫0π4sin(4t)dt is 0.

Get the solution to "∫0π4sin(4t)dt"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Celeb G.

Answered question

2022-09-23

$\int_{0}^{\frac{π}{4}} \sin (4 t) d t$

Answer & Explanation

star233

Skilled2023-05-29Added 403 answers

To solve the integral

\int_{0}^{\frac{π}{4}} \sin (4 t) d t

, we can use the basic properties of integrals and the trigonometric identity

\sin (2 θ) = 2 \sin (θ) \cos (θ)

.
Let's begin by applying the substitution

u = 4 t

. This implies

d u = 4 d t

, or equivalently,

d t = \frac{1}{4} d u

. We also need to determine the new limits of integration when we substitute

t

with

u

. When

t = 0

, we have

u = 4 \cdot 0 = 0

, and when

t = \frac{π}{4}

, we have

u = 4 \cdot \frac{π}{4} = π

. Therefore, the integral becomes:

\int_{0}^{\frac{π}{4}} \sin (4 t) d t = \frac{1}{4} \int_{0}^{π} \sin (u) d u

Next, we can evaluate the integral of

\sin (u)

with respect to

u

. Using the antiderivative of

\sin (u)

, which is

- \cos (u)

, we get:

\frac{1}{4} \int_{0}^{π} \sin (u) d u = \frac{1}{4} {[- \cos (u)]}_{0}^{π}

Now, we substitute the limits of integration:

\frac{1}{4} {[- \cos (u)]}_{0}^{π} = \frac{1}{4} [- \cos (π) - (- \cos (0))]

Since

\cos (π) = - 1

and

\cos (0) = 1

, we simplify further:

\frac{1}{4} [- \cos (π) - (- \cos (0))] = \frac{1}{4} [- (- 1) - 1]

Now, we can simplify the expression inside the brackets:

\frac{1}{4} [- (- 1) - 1] = \frac{1}{4} [1 - 1] = \frac{1}{4} \cdot 0 = 0

Therefore, the solution to the integral

\int_{0}^{\frac{π}{4}} \sin (4 t) d t

0

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?