New answers tagged trigonometric-integrals
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 ...
1
vote
Trigonometric integral with exponential
This integral can be written in the following ways:
$$
I=\begin{cases}
\int_0^\pi \exp(\cos \phi)\frac{1-\cos 2\phi}{2}d\phi&,\quad \text{substituting $\cos 2\phi=1-2\sin^2\phi$}\\
\int_0^\pi \exp(...
1
vote
Integrate $\int \frac{\mathrm dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution
If trigonometric substitution isn’t necessary, the fastest way to evaluate the given integral is to observe that it can be written as $\int\frac{\mathrm dx}{(x^2-1)\sqrt{x^2-1}}$ which is of the form $...
1
vote
Trigonometric irrational integral $ \int \frac{\sin{x}}{\cos{x}\sqrt{\cos^2{x}+\cos{x}+1}} dx $
The fastest way to evaluate the integral you were stuck at is to take $|u|$ common from the square root and perform the substitution $t=\frac1u$:
$$\begin{align}\int\frac{\mathrm du}{u\sqrt{u^2+u+1}}&...
1
vote
Evaluation of $ \int \tan x\sqrt{1+\sin x}\mathrm dx$
Alternatively, substitute $t=\sin x$:
$$\begin{align}\int\tan x\sqrt{1+\sin x}\,\mathrm dx&=\int\frac{t\sqrt{1+t}}{1-t^2}\mathrm dt\\&=\int\frac{t^2+t}{(1-t)\sqrt{1+t}}\mathrm dt\\&=-\int\...
1
vote
Accepted
Evaluate $\int{\frac{\cos x}{2-\cos x}}\mathrm dx$
To evaluate $I_1$, one way is to perform PFD:
$$\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\...
1
vote
How to reduce $\int_0^{\pi/2}\frac{1-\sin x}{\cos^2x}\sqrt{\tan x}\,dx$ to complete elliptic integral?
$$\int_0^{\pi/2}\frac{1-\sin x}{\cos^2x}\sqrt{\tan x}\,dx
\\\stackrel{t=\tan\tfrac x2}{=}\int_0^1\frac{2\sqrt2\sqrt t}{\sqrt{1-t}(1+t)^{\frac52}}dt
\\\stackrel{t=\tan^2\frac u2}{=}\frac1{\sqrt2}\int_0^...
5
votes
Evaluating $\int \sqrt{\frac{3-x}{3+x}} \sin^{-1}\left(\sqrt{\frac{3-x}{6}}\right)\mathrm dx$
Here is an alternative via integration by parts
\begin{align}
&\int\sqrt{\frac{3-x}{3+x}} \sin^{-1}\sqrt{\frac{3-x}{6}}\ dx\\
=& \int\sin^{-1}\sqrt{\frac{3-x}{6}} \bigg[\overset{ibp} d\left(\...
1
vote
Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right)\mathrm dx$
For $a\neq0$, the integrand can be simplified by using the identity
$$\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=\...
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,
$$=\int \frac{\sqrt2 \cos t -1}{2\cos^2 t-\sqrt2 \cos t+2}\,dt$$
...
2
votes
How to evaluate $\int_{-\infty}^\infty \frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}dx $
$$\begin{align}\int\frac{x-2}{(x^2+x+4)\sqrt{x^2+2x+4}}\mathrm dx&\overset{t=\frac{2-x}{2+x}}{=}4\int\frac{t\mathrm dt}{(3t^2+5)\sqrt{t^2+3}}\\&\overset{u=\sqrt{t^2+3}}{=}4\int\frac{\mathrm du}...
1
vote
How do I evaluate $\int \frac{\sqrt {x^2+16}}{x^4}dx$ using trig substitution?
An alternative method is to perform the substitution $x=\frac1{t}$(this provides an intuition to the approach of Chebyshev’s differential binomial):
$$\require{cancel}\begin{align}\int\frac{\sqrt{x^2+...
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
$$\int\frac{\...
1
vote
Evaluating $\int \frac{1}{(x^4-1)^2}dx$ without partial fraction decomposition
Consider the function
$$\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}\...
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., $t=\cot x$,
$$\begin{align}\mathcal I=\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&\overset{t=\cot x}{=}-\int\frac{\mathrm dt}{\sqrt{\...
0
votes
Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$
Multiply and divide by the conjugate of the denominator:
$$\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\\&=\...
0
votes
Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$
Performing the hyperbolic substitution $x=\tanh\theta$ since $\sqrt{1-x^2}=\text{sech}\theta$,
$$\begin{align}\int\frac{\mathrm dx}{x-\sqrt{1-x^2}}&=\int\frac{\text{sech}\theta}{\sinh\theta-1}\...
1
vote
Calculating trigonometric integral $\int_0^{\tfrac{\pi }{2}} \cos ^{-1}\frac{1}{2 \cos x+1} \, \mathrm dx$
With a few steps we can recover the integral discussed here:
$$\begin{align*}
I &= \int_0^\tfrac\pi2 \operatorname{arcsec}\left(1+2\cos x\right) \, dx \\
&= \int_0^\tfrac\pi2 \frac{x \sin x}{\...
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
$$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
0
votes
Any different way to solve $\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}dx $
Using Queen's property of definite integrals
$$\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\...
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
$$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
0
votes
Solve integral $\int\frac{1}{(x^2-1)\sqrt{x^2+1}}dx$
Breaking the $\frac1{x^2-1}$ into partial fractions, the integral becomes
$$\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\...
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
$$\mathcal I=\int_0^{2\pi}\frac{\sin x+1}{\cos x+2}\mathrm dx=2\...
0
votes
$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$ by substitution
Rewrite the denominator of the integrand as
$$(\sin x-\cos x)\cos x=\frac{2\sin x\cos x-(2\cos^2x-1)-1}2=\frac{\sin2x-\cos2x-1}2$$
Now, noting that the numerator of the integrand is $2-(\sin2x-\cos2x)^...
0
votes
How can one calculate $\int \frac{\sin^2 x}{\sin x+2\cos x}\text{d}x$?
Noting that $\sin x+2\cos x=\sqrt5\sin\left(x+\sec^{-1}\sqrt5\right)$, the integral becomes:
$$\begin{align}\int\frac{\sin^2x}{\sin x+2\cos x}\mathrm dx&\overset{t=x+\sec^{-1}\sqrt5}{=}\frac1{\...
0
votes
How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$
Substitute $t=\sin x$:
$$\begin{align}\int\frac{\mathrm dx}{\sqrt{\sin^3x+\sin^4x}}&=\text{sgn}\sin2x\int\frac{\mathrm dt}{t^\frac32(1+t)\sqrt{1-t}}\\&=\text{sgn}\sin2x\int\frac{\mathrm dt}{(t^...
0
votes
Upper and lower bounds on sine integral?
Partial answer. By $(6.2.19)$, we have
$$
\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)...
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= \...
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 $\theta$ ...
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{...
20
votes
Accepted
How to Prove This Elegant Integral Identity Involving Trigonometric and Square Root Terms
Note that
$$\frac{\left(\sqrt{1 + (a \cos t + b \sin t)^2} + a
\cos t + b \sin t\right)^2}{1 + \left(\sqrt{1 + (a \cos t + b \sin
t)^2} + a \cos t + b \sin t\right)^2} \\
=\frac12\bigg(1+\frac{a \cos ...
11
votes
How to Prove This Elegant Integral Identity Involving Trigonometric and Square Root Terms
Let $x = a\cos t + b \sin t = r \sin(t+c)$ where $r=\sqrt{a^2+b^2}$, $\sin c = a/r$, and $\cos c = b/r$. After some algebraic manipulations, we can simplify the integrand considerably:
$$
\frac{(\sqrt{...
1
vote
Accepted
Upper and lower bounds on sine integral?
We can define sharper bounds from the asymptotics.
$$\frac 1 \pi \text{Si}(\pi x)=\frac 12 +\frac {\cos(\pi x)}\pi\sum_{n=0}^\infty (-1)^{n+1}\,\frac {(2n)!}{(\pi x)^{2n+1 }}+\frac {\sin(\pi x)}\pi\...
0
votes
Upper and lower bounds on sine integral?
I'm reasonably confident that the answer is contained within the Digital Library of Mathematical Functions. The relevant section seems to be this one, although it requires a bit of cross-referencing ...
0
votes
Integrate $\int_{0}^{2\pi}\frac{1}{\sqrt{(1+a-\cos\theta)(3+a-\cos\theta)}}\mathrm{d}\theta$
Let $b=2+a$. $$\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 \...
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 ...
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}...
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 $\sin x$,...
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 $$\int\...
14
votes
How to integrate $\int\frac{dx}{\sqrt{\sin^3 x + \sin^4 x}}$
Substitute $t=\sin x$
$$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}}-{\...
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:
$$
\begin{align} I
&= \int\...
1
vote
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$
$$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}=-...
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[\...
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 ...
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 ...
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\\
&=\...
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
$...
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,
$$\int e^{ax} \cos(bx+c)\,dx=\frac{\mathrm{e}^{ax}}{a^2+b^2}{ \left(a \cos\left(bx + c\right)+b \sin\...
1
vote
What's wrong with my approach in evaluating $\int\frac{\mathrm dx}{(a^2x^2-b^2)^\frac32}$ via trigonometric substitution?
You would be on the right track had the integral been
$$\int\frac{\mathrm dx}{(a^2x^2\color{red}{+}b^2)^\frac32}$$
Even if the integral was this, you have made a few mistakes:
$1.$ You have forgotten ...
1
vote
Evaluate $\int e^{-3x} \cos^3x\,dx$
I’ll give an answer that is completely free of any binomial or IBP tediousness, that I personally think is cleaner and faster. First, use the identity
$$\cos3x=4\cos^3x-3\cos x$$
to rewrite the ...
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