Integral Problems & Solutions - Explore with Us!

Recent questions in Applications of integrals

Ashley Fritz 2022-04-10

Compute $\int \frac{8 \tan^{- 1} (4 x)^{9}}{3 (16 x^{2} + 1)} d x$ .

Ainsley Zimmerman 2022-04-10

Integrate $\frac{4 e^{\tan^{- 1} (t)}}{7 (t^{2} + 1)}$ with respect to t.

Micah Haynes 2022-04-10

Evaluate $\int \frac{4}{9} e^{6 x} \tan (e^{6 x} + 4) \sec (e^{6 x} + 4) d x$ .

Noelle Wright 2022-04-10

Evaluate $\int \frac{8}{3} \tan (4 t) e^{\sec (4 t)} \sec (4 t) d t$ .

Tiffany Maldonado 2022-04-10

How do you find the primitive/integral of $\arctan x$ ?
I tried searching for how you derive the integral/primitive of $\arctan (x)$ , but I can't find any question on S.E with an answer that clearly explains this. I feel that there should be one, since it is rather fundamental knowledge, and it does come up every now and then as part of S.E questions.
So the primitive can be looked up and it should be the following:
$\int \arctan (x) d x = x \arctan (x) - \frac{1}{2} \ln (1 + x^{2}) + C$
How do you get the result?

Ormezzani6cuu 2022-04-09

trigonometric limit with integral: $lim_{α \to 0} \int_{0}^{α} \frac{d x}{\sqrt{\cos x - \cos α}}$

Karla Thompson 2022-04-07

How to integrate $\tan^{- 1} (\frac{1}{2 \sin (x)})$
I want to calculate the following integral :
$\int_{{0}}^{\frac{π}{2}} \tan^{- 1} (\frac{1}{2 \sin (x)}) d x$
But I don't how; I tried by subsituting $u = \frac{1}{2 \sin (x)}$ and $u = \tan^{- 1} (\frac{1}{2 \sin (x)})$ but it doesn't lead me anywhere.

Jaylen Cantrell 2022-04-06

Find value of following integral $\int \frac{x \tan (x) \sec (x)}{{(\tan (x) - x)}^{2}} d x$
the numerator is $d [\sec (x)]$ but that isnt work due to x in denominator. First we can simplify as
$\int \frac{x \sin (x)}{{(\sin (x) - x \cos (x))}^{2}} dx = \int \frac{x}{(\sin (x) - x \cos (x))} dx + \int \frac{x^{2} \cos (x)}{{(\sin (x) - x \cos (x))}^{2}} dx$
but again its not manipulative. Suggest a useful substitution or method.

Heather Dickson 2022-04-06

I want to compute
$\int_{- \infty}^{\infty} \frac{1}{\sqrt{x + y i + 2}} d y$
where i is the imaginary number. How to compute this integral?

peggyleuwpodg 2022-04-05

Evaulating the trigonometric integral $\int \frac{1}{{(x^{2} + 1)}^{2}} d x$
To do this, I let $x = \tan u$ . Now we have $d x = \sec^{2} u d u$
$\int \frac{1}{{(x^{2} + 1)}^{2}} d x = \int \frac{\sec^{2} {u} d u}{{(\tan^{2} {u} + 1)}^{2}}$
$\int \frac{1}{{(x^{2} + 1)}^{2}} d x = \int \frac{1}{\sec^{2} {u}} d u = \int \cos^{2} {u} d u$
$\int \frac{1}{{(x^{2} + 1)}^{2}} d x = \int \frac{\cos {(2 u)} + 1}{2} d u = \frac{\sin (u)}{4} + \frac{u}{2}$
$\int \frac{1}{{(x^{2} + 1)}^{2}} d x = \frac{\sqrt{1 - \cos^{2} {u}}}{4} + \frac{u}{2}$
Now, I think I am right so far but I do not know have to get rid of the u in the $\cos^{2} (u)$ term. Please help.

Erik Cantu 2022-04-02

Existence of the integral $\int_{α}^{β} \sqrt{\frac{\arctan h (r)}{r}} d r$
For $- 1 \leq α \leq β \leq 1$

Using the integral test, find the positive values of p for which the series ∑∞ k=2 1/ (k(ln(k)))^p converges. Show your work and explain your reasoning.

Rolando Wade 2022-04-01

I have the equation:
$\frac{1}{τ} \int_{0}^{τ} A \sin (Ω t) \cdot A \sin (Ω (t - λ)) d t$
for which the attempted solution is to convert the sine terms into complex natural exponents (engineering notation using j as imaginary unit) as
$\frac{A^{2}}{τ} \int_{0}^{τ} \frac{e^{j Ω t} - e^{- j Ω t}}{2 j} \cdot \frac{e^{j Ω (t - λ)} - e^{- j Ω (t - λ)}}{2 j} d t$
the next step in the solution moves the $\frac{1}{2 j}$ term outside of the integral to form
$\frac{- A^{2}}{4 τ} \int_{0}^{τ} (e^{j Ω t} - e^{- j Ω t}) \cdot (e^{j Ω (t - λ)} - e^{- j Ω (t - λ)}) d t$
I'm struggling to understand how $\frac{1}{2 j} \to \frac{- 1}{4}$ when being moved out of the integral.

Oxinailelpels3t14 2022-04-01

I am trying to evaluate the integral
$\int_{0}^{\frac{π}{2}} \frac{\log (e^{i x} + e^{- i x})] - \log 2}{e^{2 i x} - 1} e^{i x} d x$
I am wondering if a complex variable substitution the likes of $i u = e^{i x}$ , $d u = e^{i x} d x$ is justifiable? In my mind. when x=0,u=0 and when $x = \frac{π}{2}, u = 1$ which would make the integral
$- \int_{0}^{1} \frac{\log (i u - \frac{i}{u}) - \log (2)}{1 + u^{2}}, d u$

Averie Ferguson 2022-03-31

Solving periodic equation $\cos 7 θ = \cos 3 θ + \sin 5 θ$
I approached the problem as follows : $\cos 7 θ = \cos 3 θ + \sin 5 θ \Rightarrow \cos 7 θ - \cos 3 θ = \sin 5 θ \Rightarrow 2 \sin 5 θ \sin (- 2 θ) = \sin 5 θ$
From there on we get, $\sin 2 θ = - \frac{1}{2} and \sin 5 θ = 0$
So, we deduce that $θ = - \frac{π}{12} + π k$ and $θ = \frac{2}{5} π n$ where k and n are integers.
However, Wolfram|Alpha doesn't seem to agree with me. They present $\frac{π}{5}$ as a solution which is not included in my solution.

Lucian Ayers 2022-03-31

$\int \frac{d x}{1 + x}, x \geq 0$ by trigonometric substitution
Consider the integral $I = \int \frac{d x}{1 + x}, x \in [0, \infty)$ . A standard treatment using the substitution u=1+x directly gives the result $\ln (1 + x) + c$ . Now consider doing this with the trigonometric substitution $\sqrt{x} = \tan θ, θ \in [0, \frac{π}{2})$ then $x = \tan^{2} θ, d x = 2 \tan θ \sec^{2} θ$ . Now, following your nose with right triangle trigonometry this leads directly to the solution $I = 2 \ln (1 + x) + c$ .

svezic0rn 2022-03-31

Proving the double differential of $z = - z$ implies $z = \sin x$
$\frac{d^{2} z}{{d x}^{2}} = - z$
implies z is of the form $a \sin x + b \cos x$ . Is there a proof for the same. I was trying to arrive at the desired function but couldn't understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of $\sin x or \cos x$ ?

dirafe7455 2022-03-29

A particle moves along a line with velocity function v(t)=t^2-t, where v is measured in meters per second. Find (a) the displacement and (b) the distance traveled by the particle during the time interval [0,5].

blestimd4pz 2022-03-29

How to solve $\int \sin^{3} (x) \cos^{2} (x) d x$ with integration by parts?

ikramkeyslo4s 2022-03-27

How Can you factor 1/2 out of integral of $\cos 2 x$ ?
$\int \cos 2 x d x$
simplifies to
$\frac{1}{2} \int \cos u d u$
where u=2x
My Question is where does the $\frac{1}{2}$ that is factored out come from?
It would make more sense for a factor of 1/2 to be extracted if the integral was:
$\int \frac{\cos x}{2} d x$

Applications of integrals are one of the popular but still complex fields that many students encounter not only in Math and Physics but in subjects like Programming and Engineering. For example, start with the applications of double integrals and you will see how it is useful in the construction of some building where the shape must be found or the gravity matters should be presented as the model. At the same time, if your questions are based on equations, consider geometric applications of definite integrals help that is also present below as you take a look through the relevant answers.

Integral Calculus

Master Integral Application Problems