Skip to main content

New answers tagged

1 vote

Need help with $\int_0^\pi\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$

Here is an elementary solution. Consider the generalized integral below \begin{align} I(a)=&\int_0^\pi\arctan^2 \frac{a\sin x}{1+a\cos x}\ dx\\ I’(a)=& \int_0^\pi\arctan\frac{a\sin x}{1+a\cos ...
Quanto's user avatar
  • 108k
1 vote

Trigonometric integral with exponential

This integral can be written in the following ways: $$ I=\begin{cases} \int_0^\pi \exp(\cos \phi)\frac{1-\cos 2\phi}{2}d\phi&,\quad \text{substituting $\cos 2\phi=1-2\sin^2\phi$}\\ \int_0^\pi \exp(...
Mostafa Ayaz's user avatar
  • 32.8k
1 vote

Integrate $\int \frac{\mathrm dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution

If trigonometric substitution isn’t necessary, the fastest way to evaluate the given integral is to observe that it can be written as $\int\frac{\mathrm dx}{(x^2-1)\sqrt{x^2-1}}$ which is of the form $...
Math Guy's user avatar
  • 4,488
1 vote

Trigonometric irrational integral $ \int \frac{\sin{x}}{\cos{x}\sqrt{\cos^2{x}+\cos{x}+1}} dx $

The fastest way to evaluate the integral you were stuck at is to take $|u|$ common from the square root and perform the substitution $t=\frac1u$: $$\begin{align}\int\frac{\mathrm du}{u\sqrt{u^2+u+1}}&...
Math Guy's user avatar
  • 4,488
1 vote

Evaluation of $ \int \tan x\sqrt{1+\sin x}\mathrm dx$

Alternatively, substitute $t=\sin x$: $$\begin{align}\int\tan x\sqrt{1+\sin x}\,\mathrm dx&=\int\frac{t\sqrt{1+t}}{1-t^2}\mathrm dt\\&=\int\frac{t^2+t}{(1-t)\sqrt{1+t}}\mathrm dt\\&=-\int\...
Math Guy's user avatar
  • 4,488
1 vote
Accepted

Evaluate $\int{\frac{\cos x}{2-\cos x}}\mathrm dx$

To evaluate $I_1$, one way is to perform PFD: $$\begin{align}I_1&=\int\frac{2\sec^2\frac{x}2}{(3\tan^2\frac{x}2+1)(\tan^2\frac{x}2+1)}\mathrm dx\\&\overset{t=\tan\frac{x}2}{=}\int\frac{4\...
Math Guy's user avatar
  • 4,488
1 vote

How to reduce $\int_0^{\pi/2}\frac{1-\sin x}{\cos^2x}\sqrt{\tan x}\,dx$ to complete elliptic integral?

$$\int_0^{\pi/2}\frac{1-\sin x}{\cos^2x}\sqrt{\tan x}\,dx \\\stackrel{t=\tan\tfrac x2}{=}\int_0^1\frac{2\sqrt2\sqrt t}{\sqrt{1-t}(1+t)^{\frac52}}dt \\\stackrel{t=\tan^2\frac u2}{=}\frac1{\sqrt2}\int_0^...
Bob Dobbs's user avatar
  • 14k
5 votes

Evaluating $\int \sqrt{\frac{3-x}{3+x}} \sin^{-1}\left(\sqrt{\frac{3-x}{6}}\right)\mathrm dx$

Here is an alternative via integration by parts \begin{align} &\int\sqrt{\frac{3-x}{3+x}} \sin^{-1}\sqrt{\frac{3-x}{6}}\ dx\\ =& \int\sin^{-1}\sqrt{\frac{3-x}{6}} \bigg[\overset{ibp} d\left(\...
Quanto's user avatar
  • 108k
1 vote

Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right)\mathrm dx$

For $a\neq0$, the integrand can be simplified by using the identity $$\sin^{-1}x=\tan^{-1}\frac{x}{\sqrt{1-x^2}}\forall x\in[0,1)$$ $$\implies \mathcal I=\int\sin^{-1}\sqrt{\frac{x}{x+a}}\mathrm dx=\...
Math Guy's user avatar
  • 4,488
0 votes

Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$

Consider, $$\int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}$$ Substitute $x-\frac{\pi}{4}=t$ After a lot of simplification, $$=\int \frac{\sqrt2 \cos t -1}{2\cos^2 t-\sqrt2 \cos t+2}\,dt$$ ...
Amrut Ayan's user avatar
  • 5,701
2 votes

How to evaluate $\int_{-\infty}^\infty \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}dx $

$$\begin{align}\int\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx&\overset{t=\frac{2-x}{2+x}}{=}4\int\frac{t\mathrm dt}{(3t^2+5)\sqrt{t^2+3}}\\&\overset{u=\sqrt{t^2+3}}{=}4\int\frac{\mathrm du}...
Math Guy's user avatar
  • 4,488
1 vote

How do I evaluate $\int \frac{\sqrt {x^2+16}}{x^4}dx$ using trig substitution?

An alternative method is to perform the substitution $x=\frac1{t}$(this provides an intuition to the approach of Chebyshev’s differential binomial): $$\require{cancel}\begin{align}\int\frac{\sqrt{x^2+...
Math Guy's user avatar
  • 4,488
0 votes

Evaluate $\int\frac{x^2}{(3+4x-4x^2)^{3/2}}dx$

A clean way to avoid trigonometric substitution is to break the numerator of the integrand as follows: $$x^2=-\frac14(-4x^2+4x+3)-\frac18(-8x+8)+\frac74$$ Now, just use the results $$\int\frac{\...
Math Guy's user avatar
  • 4,488
1 vote

Evaluating $\int \frac{1}{(x^4-1)^2}dx$ without partial fraction decomposition

Consider the function $$\mathcal I(a)=\int_0^x\frac{\mathrm dt}{t^4-a^4}\tag{1}$$ Since $(t^4-a^4)=(t^2+a^2)(t^2-a^2)$, another expression for $\mathcal I(a)$ is $$\mathcal I(a)=\frac1{2a^3}\tan^{-1}\...
Math Guy's user avatar
  • 4,488
0 votes

How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

Going by the first substitution that comes to mind, i.e., $t=\cot x$, $$\begin{align}\mathcal I=\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&\overset{t=\cot x}{=}-\int\frac{\mathrm dt}{\sqrt{\...
Math Guy's user avatar
  • 4,488
0 votes

Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$

Multiply and divide by the conjugate of the denominator: $$\begin{align}\int\frac{\mathrm dx}{x-\sqrt{1-x^2}}&=\int\frac{x}{2x^2-1}\mathrm dx+\int\frac{\sqrt{1-x^2}}{2x^2-1}\mathrm dx\\&=\...
Math Guy's user avatar
  • 4,488
0 votes

Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$

Performing the hyperbolic substitution $x=\tanh\theta$ since $\sqrt{1-x^2}=\text{sech}\theta$, $$\begin{align}\int\frac{\mathrm dx}{x-\sqrt{1-x^2}}&=\int\frac{\text{sech}\theta}{\sinh\theta-1}\...
Math Guy's user avatar
  • 4,488
1 vote

Calculating trigonometric integral $\int_0^{\tfrac{\pi }{2}} \cos ^{-1}\frac{1}{2 \cos x+1} \, \mathrm dx$

With a few steps we can recover the integral discussed here: $$\begin{align*} I &= \int_0^\tfrac\pi2 \operatorname{arcsec}\left(1+2\cos x\right) \, dx \\ &= \int_0^\tfrac\pi2 \frac{x \sin x}{\...
user170231's user avatar
  • 22.5k
0 votes

Any different way to solve $\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}dx $

Using Queen's property of definite integrals $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ the given integral becomes $$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
Math Guy's user avatar
  • 4,488
0 votes

Any different way to solve $\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}dx $

Using Queen's property of definite integrals $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ the given integral becomes $$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
Math Guy's user avatar
  • 4,488
0 votes

Any different way to solve $\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}dx $

Using Queen's property of definite integrals $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ the given integral becomes $$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
Math Guy's user avatar
  • 4,488
0 votes

Solve integral $\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$

Breaking the $\frac1{x^2-1}$ into partial fractions, the integral becomes $$\begin{align}\int\frac{\mathrm dx}{(x^2-1)\sqrt{x^2+1}}&=\frac12\int\frac{\mathrm dx}{(x-1)\sqrt{x^2+1}}-\frac12\int\...
Math Guy's user avatar
  • 4,488
1 vote

Any different way to solve $\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}dx $

Using Queen's property of definite integrals $$\int_0^{2a}f(x)\mathrm dx=\int_0^a f(x)+f(2a-x)\mathrm dx$$ the given integral becomes $$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
Math Guy's user avatar
  • 4,488
0 votes

$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$ by substitution

Rewrite the denominator of the integrand as $$(\sin x-\cos x)\cos x=\frac{2\sin x\cos x-(2\cos^2x-1)-1}2=\frac{\sin2x-\cos2x-1}2$$ Now, noting that the numerator of the integrand is $2-(\sin2x-\cos2x)^...
Math Guy's user avatar
  • 4,488
0 votes

How can one calculate $\int \frac{\sin^2 x}{\sin x+2\cos x}\text{d}x$?

Noting that $\sin x+2\cos x=\sqrt5\sin\left(x+\sec^{-1}\sqrt5\right)$, the integral becomes: $$\begin{align}\int\frac{\sin^2x}{\sin x+2\cos x}\mathrm dx&\overset{t=x+\sec^{-1}\sqrt5}{=}\frac1{\...
Math Guy's user avatar
  • 4,488
0 votes

How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

Substitute $t=\sin x$: $$\begin{align}\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&=\text{sgn}\sin2x\int\frac{\mathrm dt}{t^\frac32(1+t)\sqrt{1-t}}\\&=\text{sgn}\sin2x\int\frac{\mathrm dt}{(t^...
Math Guy's user avatar
  • 4,488
0 votes

Upper and lower bounds on sine integral?

Partial answer. By $(6.2.19)$, we have $$ \frac{1}{\pi }\operatorname{Si}(\pi x) = \frac{1}{2} - \frac{1}{\pi }\left( {\operatorname{f}(\pi x)\cos (\pi x) + \operatorname{g}(\pi x)\sin (\pi x)} \right)...
Gary's user avatar
  • 34.2k
1 vote

Compute $ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x) d \theta}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} }$

Differentiate $g(x)$ below to obtain $f(x)$ \begin{align} g(x) =& \int_0^{\pi} \frac{ \sin \theta}{ \sqrt{x^2-2x\cos\theta+1 }}d\theta\\ =&\ \frac1x \sqrt{x^2-2x\cos\theta+1}\bigg|_0^\pi= \...
Quanto's user avatar
  • 108k
1 vote

$\sqrt{t^2}$ and $\sin^{-1}(\sin x)$ in $\int \sqrt{\frac{3-x}{3+x}}\sin^{-1}\left(\sqrt{\frac{3-x}6}\right)\mathrm dx$

I figured out that my solution is incomplete and my teacher’s final answer is correct, although the details(of a bijective substitution) were overlooked. I missed out stating the domain of $\theta$ ...
Math Guy's user avatar
  • 4,488
4 votes

Integrals of trignometric functions $\cot^{-1}\sqrt{1+\csc{\theta}}$ and $\csc^{-1}\sqrt{1+\cot{\theta}}$

For the first integral, substitute $\csc\theta=\frac{1+\sin^2t}{1-\sin^2t}$ \begin{align}\int_0^{{\pi}/{2}}\cot^{-1} &{\sqrt{1+\csc{\theta}}} \ {d\theta} =\int_0^{\pi/2} \int_0^{\pi/4}\frac{\sqrt{...
Quanto's user avatar
  • 108k
20 votes
Accepted

How to Prove This Elegant Integral Identity Involving Trigonometric and Square Root Terms

Note that $$\frac{\left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2}{1 + \left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2} \\ =\frac12\bigg(1+\frac{a \cos ...
Quanto's user avatar
  • 108k
11 votes

How to Prove This Elegant Integral Identity Involving Trigonometric and Square Root Terms

Let $x = a\cos t + b \sin t = r \sin(t+c)$ where $r=\sqrt{a^2+b^2}$, $\sin c = a/r$, and $\cos c = b/r$. After some algebraic manipulations, we can simplify the integrand considerably: $$ \frac{(\sqrt{...
Pranay's user avatar
  • 3,693
1 vote
Accepted

Upper and lower bounds on sine integral?

We can define sharper bounds from the asymptotics. $$\frac 1 \pi \text{Si}(\pi x)=\frac 12 +\frac {\cos(\pi x)}\pi\sum_{n=0}^\infty (-1)^{n+1}\,\frac {(2n)!}{(\pi x)^{2n+1 }}+\frac {\sin(\pi x)}\pi\...
Claude Leibovici's user avatar
0 votes

Upper and lower bounds on sine integral?

I'm reasonably confident that the answer is contained within the Digital Library of Mathematical Functions. The relevant section seems to be this one, although it requires a bit of cross-referencing ...
Greg Martin's user avatar
  • 86.9k
0 votes

Integrate $\int_{0}^{2\pi}\frac{1}{\sqrt{(1+a-\cos\theta)(3+a-\cos\theta)}}\mathrm{d}\theta$

Let $b=2+a$. $$\begin{align*} I &= \int_0^{2\pi} \frac{dt}{\sqrt{(b-1-\cos t)(b+1-\cos t)}} \\ &= 2\int_0^\pi \frac{dt}{\sqrt{(b-1+\cos t)(b+1+\cos t)}} & t\to t+\pi \\ &= \frac4b \...
user170231's user avatar
  • 22.5k
1 vote

Why does integrating with polar coordinates give different answers depending on if $\frac{3\pi}{2}$ or $-\frac{\pi}{2}$ is used?

We say the line $\theta = \frac{3\pi}{2}$ is equivalent to $\theta = -\frac{\pi}{2}$, because they are $2\pi$ apart, and if we go $2\pi$ radians around the origin, we are where we started. So the ...
altwoa's user avatar
  • 159
2 votes
Accepted

General solution for $ \int \frac{dx}{(x^2 + a^2)^2} $

Hint: Substitute $x=\sqrt{a}\tan(\theta)$ Now, the given integral is $$\int\frac{1}{(x^{2}+a)^{2}}\mathrm dx$$ Now, if I substitute $x=\sqrt{a}\tan(\theta)$ then I will get $\mathrm dx=\sqrt{a}\sec^{2}...
Mathematics's user avatar
10 votes

How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

By periodicity of the integrand and positivity of the argument of the radical, we may as well assume we are working on the interval $(0, \pi)$. Since the integrand is an algebraic function in $\sin x$,...
Travis Willse's user avatar
6 votes
Accepted

How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

The given integral is $$\int\frac{1}{\sqrt{\sin^{3}(x)+\sin^{4}(x)}}\mathrm dx$$ Trick: Take $\sin^{4}(x)$ common from the square root in the denominator. Therefore the integral will become $$\int\...
Mathematics's user avatar
14 votes

How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$

Substitute $t=\sin x$ $$I= \int \frac1{t^{3/2}(1+t)\sqrt{1-t}}dt $$ Then, $1-t=y^2$ $$I= -\int \frac2{(1-y^2)^{3/2}}-\frac2{(2-y^2) \sqrt{1-y^2}} \ dy $$ As a result $$I= \frac{2y}{\sqrt{1-y^2}}-{\...
Quanto's user avatar
  • 108k
6 votes
Accepted

Solving $\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sec x}+\sqrt{\csc x}}dx$

Recall Equation $(3.681\,\text{-}\,1)$ from Book Table of Integrals, Series, and Products by I.S. Gradshtein and I.M. Ryzhik, Page $(411)$: Simplify the Integral: $$ \begin{align} I &= \int\...
Hazem Orabi's user avatar
  • 4,249
1 vote

How to evaluate $\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$

$$I=\int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x}\,dx$$ Re-writing the integral, $$I=-\frac{1}{2}\int_0^{\frac{\pi}{2}} \frac{\ln (1+\tan^2x)}{1+2\tan^2x}\sec^2x\,dx$$ $$I\overset{\tan x=x}=-...
Amrut Ayan's user avatar
  • 5,701
1 vote

Solving $\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sec x}+\sqrt{\csc x}}dx$

Another way ,using Mellin Transform with Inverse Mellin transform: $\int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sec (x)}+\sqrt{\csc (x)}} \, dx=\\\underset{A\to 1}{\text{lim}}\mathcal{M}_s^{-1}\left[\...
Mariusz Iwaniuk's user avatar
0 votes

How do I evaluate $\int \frac{x^4}{\sqrt{4-x^2}}dx$ using trig substitution?

You've missed a factor of $2$ while performing the substitution $x=2\sin\theta$, so the integrand is actually $16\sin^4\theta$, not $8\sin^4\theta$. Alternatively, you can finish off the integration ...
Math Guy's user avatar
  • 4,488
1 vote

How do I evaluate $\int \frac{\sqrt {x^2+16}}{x^4}dx$ using trig substitution?

Alternatively, note that the integral is of the form of Chebyshev's differential binomial and belongs to the case $\frac{m+1}{n}+p\in\mathbb Z$(where $m=-4$, $n=2$ and $p=\frac12$). Hence, we perform ...
Math Guy's user avatar
  • 4,488
6 votes

Closed form for integral of inverse hyperbolic function $\int_{0}^{\frac{\pi}{2}}\sinh^{-1}{\left(\sqrt{\sin{x}}\right)}\,\mathrm{d}x$

Substitute $\sin x= \cos 2y$ to have $dx =-2 dy$ \begin{align} \int_{0}^{{\pi}/{2}} &\sinh^{-1} \left(\sqrt{\sin{x}}\right)dx = \int_0^{\pi/4}2 \cosh^{-1}\left(\sqrt2\cos y\right) dy\\ &=\...
Quanto's user avatar
  • 108k
2 votes

Evaluate $\int e^{-3x} \cos^3x\,dx$

Mechanically reduction formulas or complex approach is good. But, most of the time I prefer the method of undetermined coefficients. It is practical for small degrees. Using Alberto Andrenucci's hint $...
Bob Dobbs's user avatar
  • 14k
3 votes

Evaluate $\int e^{-3x} \cos^3x\,dx$

After converting the cosine term into a sum of linear cosine terms you can use the below formulae, $$\int e^{ax} \cos(bx+c)\,dx=\frac{\mathrm{e}^{ax}}{a^2+b^2}{ \left(a \cos\left(bx + c\right)+b \sin\...
Amrut Ayan's user avatar
  • 5,701
1 vote

What's wrong with my approach in evaluating $\int\frac{\mathrm dx}{(a^2x^2-b^2)^\frac32}$ via trigonometric substitution?

You would be on the right track had the integral been $$\int\frac{\mathrm dx}{(a^2x^2\color{red}{+}b^2)^\frac32}$$ Even if the integral was this, you have made a few mistakes: $1.$ You have forgotten ...
Math Guy's user avatar
  • 4,488
1 vote

Evaluate $\int e^{-3x} \cos^3x\,dx$

I’ll give an answer that is completely free of any binomial or IBP tediousness, that I personally think is cleaner and faster. First, use the identity $$\cos3x=4\cos^3x-3\cos x$$ to rewrite the ...
Math Guy's user avatar
  • 4,488

Top 50 recent answers are included