Some indefinite integrals

Calculus (Integrals, Series)
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Tsakanikas Nickos
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Some indefinite integrals

#1

Post by Tsakanikas Nickos »

Compute the following indefinite integrals

(i) \( \displaystyle \int \frac{ \tan{x} }{ \sqrt{ a+b\tan^{2}{x} } } \mathrm{d}x, \; b>a \)

(ii) \( \displaystyle \int \frac{ \sin{x} }{ \sqrt{ 2 + \sin(2x) } } \mathrm{d}x \)

(iii) \( \displaystyle \int \frac{1}{x\sqrt[3]{x^{2}+1}} \mathrm{d}x \)
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Grigorios Kostakos
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Re: Some indefinite integrals

#2

Post by Grigorios Kostakos »

We give calculations for the integrals \(i.\) and \(iii.\) :

\(i.\) \begin{align*}
\displaystyle \int{\frac{ \tan{x} }{ \sqrt{ a+b\tan^{2}{x} } } \mathrm{d}x}&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\tan^2{x}}\\
{\frac{1}{2(1+t)}\,dt\,=\,\tan{x}\,dx}\\
\end{subarray}}\, \int{\frac{1}{2(1+t)\sqrt{a+bt}} \mathrm{d}t}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,\sqrt{a+bt}}\\
{\frac{1}{b}\,du\,=\,\frac{1}{2\sqrt{a+bt}}\,dt}\\
\end{subarray}}\,\int{\frac{1}{u^2+b-a} \mathrm{d}t}\\
&=\frac{1}{b-a}\int{\frac{}{\bigl({\frac{u}{\sqrt{b-a}}}\bigr)^2+1} \mathrm{d}t}\\
&=\frac{1}{\sqrt{b-a}}\int{\frac{1}{\bigl({\frac{u}{\sqrt{b-a}}}\bigr)^2+1} \mathrm{d}\bigl({\tfrac{u}{\sqrt{b-a}}}\bigr)}\\
&=\frac{1}{\sqrt{b-a}}\arctan\bigl({\tfrac{u}{\sqrt{b-a}}}\bigr)+c\\
&\stackrel{u\,=\,\sqrt{a+b\tan^2{x}}}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\frac{1}{\sqrt{b-a}}\arctan\Bigl({\sqrt{\tfrac{a+b\tan^2{x}}{b-a}}\,}\Bigr)+c\,, \quad b>a\,.
\end{align*}

\(iii.\) \begin{align*}
\displaystyle \int{\frac{1}{x\sqrt[3]{x^{2}+1}} \mathrm{d}x}&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t^3\,=\,1+x^2}\\
{\frac{3t^2}{2\sqrt{t^3-1}}\,dt\,=\,dx}\\
\end{subarray}}\,\int{\frac{1}{t\sqrt{t^3-1}}\frac{3t^2}{2\sqrt{t^3-1}}\mathrm{d}t}\\
&=\frac{1}{2}\int{\frac{3t}{t^3-1}\mathrm{d}t}\\
&=\frac{1}{2}\int{\frac{1}{t-1}\mathrm{d}t}+\frac{1}{2}\int{\frac{-t+1}{t^2+t+1}\mathrm{d}t}\\
&=\frac{1}{2}\log|t-1|+c_1+\frac{1}{4}\int{\frac{-2t-1}{t^2+t+1}\mathrm{d}t}+\frac{1}{4}\int{\frac{3}{t^2+t+1}\mathrm{d}t}\\
&=\frac{1}{2}\log|t-1|+c_1-\frac{1}{4}\int{\frac{(t^2+t+1)'}{t^2+t+1}\mathrm{d}t}+\int{\frac{3}{(2t+1)^2+3}\mathrm{d}t}\\
&=\frac{1}{2}\log|t-1|-\frac{1}{4}\log(t^2+t+1)+c_2+\frac{\sqrt{3}}{2}\int{\frac{1}{\bigl({\frac{2t+1}{\sqrt{3}}}\bigr)^2+1}\mathrm{d}\bigl({\tfrac{2t+1}{\sqrt{3}}}\bigr)}\\
&=\frac{1}{2}\log|t-1|-\frac{1}{4}\log(t^2+t+1)+\frac{\sqrt{3}}{2}\arctan\bigl({\tfrac{2t+1}{\sqrt{3}}}\bigr)+c\\
&=\frac{1}{4}\log\bigl({\tfrac{t^2-2t+1}{t^2+t+1}}\bigr)+\frac{\sqrt{3}}{2}\arctan\bigl({\tfrac{2t+1}{\sqrt{3}}}\bigr)+c\\
&\stackrel{t\,=\,\sqrt[3]{1+x^2}}{=\!=\!=\!=\!=\!=}\frac{1}{4}\log\Bigl({\tfrac{{\sqrt[3]{1+x^2}\,}^2-2\sqrt[3]{1+x^2}+1}{{\sqrt[3]{1+x^2}\,}^2+\sqrt[3]{1+x^2}+1}}\Bigr)+\frac{\sqrt{3}}{2}\arctan\bigl({\tfrac{2\sqrt[3]{1+x^2}+1}{\sqrt{3}}}\bigr)+c \,.
\end{align*}
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Tolaso J Kos
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Re: Some indefinite integrals

#3

Post by Tolaso J Kos »

Good evening Nickos...
Let me answer, for now, integrals \( ii. \,\,\,\, iii. \).

\(ii. \)
$$\begin{aligned}
\int \frac{\sin x}{\sqrt{1+\sin 2x}}\, dx &= \int \frac{\sin x\sqrt{1+\sin 2x}}{\left ( \sin x+\cos x \right )^2}\, dx \\
&\overset{\left ( \ast \right )}{=\!} \int \frac{\tan x\sec^2 x}{1+\tan x+\tan^2 x+\tan^3 x}\, dx\\
&\overset{u=\tan x}{=\! =\! =\! =\!}\int \frac{u}{1+u+u^2+u^3}\, du \\
&=\int \left ( \frac{u+1}{2\left ( u^2+1 \right )}-\frac{1}{2(u+1)} \right )\, du \\
&= \frac{1}{2}\int \left ( \frac{u}{u^2+1}+\frac{1}{u^2+1} \right )\, du-\frac{1}{2}\int \frac{du}{u+1} \\
&=\frac{1}{4}\ln\left ( u^2+1 \right )+\frac{1}{2}\tan^{-1}u-\frac{1}{2}\ln\left | u+1 \right |
\end{aligned}$$

Now in terms of \( x \) the result is: \( \displaystyle \frac{x}{2}-\frac{1}{2}\ln \left | \tan x+1 \right |+\frac{1}{4}\ln \left ( \sec^2 x \right )+c, \,\,\, c\in \mathbb{R} \).

\(iii. \)
Well, this had me a little stumbed. For starters I applied the substitution \( x=\tan u \) however it didn't work.
Then I tried the sub \( x^2+1=u \) and I got:

$$\begin{aligned}
\int \frac{dx}{x\sqrt[3]{x^2+1}} &\overset{u=x^2+1}{=\! =\! =\! =\!}\frac{1}{2}\int\frac{du}{\sqrt{u-1}\sqrt{u-1}\sqrt[3]{u}} \\
&= \frac{1}{2}\int \frac{du}{(u-1)\sqrt[3]{u}}\\
&\overset{y=\sqrt[3]{u}}{=\! =\! =\!}\frac{3}{2}\int \frac{y}{y^3-1}dy \\
&= \frac{3}{2}\int \left ( \frac{1-y}{3(y^2+y+1)}+\frac{1}{3(y-1)} \right )dy\\
&= \frac{1}{2}\int \frac{1-y}{y^2+y+1}dy+\frac{1}{2}\int \frac{dy}{y-1} \\
&=\cdots
\end{aligned}$$

In order to integrate the integral: \( \displaystyle J=\int \frac{1-y}{y^2+y+1}\, dy \) since we cannot analyse it in partial functions (the denominator is not factored in \( \mathbb{R} \)) we can apply a trick: We can rewritte as: \( \displaystyle J=\int \frac{1-y}{y^2+y+1}\, dy=\int \frac{3-2y-1}{2\left ( y^2+y+1 \right )}dy=\int \left ( \frac{3}{2\left ( y^2+y+1 \right )}-\frac{2y+1}{2\left ( y^2+y+1 \right )} \right ) \, dy=\cdots \) and continue for the first integral using a direct substituiton of the denominator and for the second integral ohh... we just see that the nominator is simply the derivative of the denominator thus we quickly get an \( \ln \) out of that and then it is a matter of operations that I ommit them because it's a long shot till the very end.

\( (*) \) I multiplied nominator and denominator by \( \displaystyle \frac{\sec^4 x}{\tan x+1} \) and after some operations I got to the equation on the right.

I'll be back for the first one.
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Tsakanikas Nickos
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Re: Some indefinite integrals

#4

Post by Tsakanikas Nickos »

Thank you, Tolis, for your solutions! I'm looking forward to seeing your solution to the first integral, which, i believe, is the most interesting of the three and probably the most difficult.

Let me point out some things about the third integral:

(1) A solution i've seen is based exactly on the substitution \( \displaystyle x = \tan{u}. \) The result is a function of \( \displaystyle t = \sqrt[3]{x^2+1}, \) if i'm not mistaken. Maybe you'd like to try it again this way and see whether it actually works out!

(2) Indeed the denominator of J doesn't factor over \( \mathbb{R} \). However, there is another way to compute it. Completing the squares of the demoninator, we get
\[ \displaystyle y^2 + y + 1 = \left( y+\frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} \]
and it's easily seen that
\begin{align*}
\displaystyle
J &= \int \frac{1-y}{ \left( y+\frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} } \mathrm{d}y \\
&= \left( \frac{2}{\sqrt{3}} \right)^{2} \left[ \int \frac{\mathrm{d}y}{\left( \frac{2y+1}{\sqrt{3}} \right)^{2} + 1} - \int \frac{y\mathrm{d}y}{\left( \frac{2y+1}{\sqrt{3}} \right)^{2} + 1} \right] \\
&\overset{(*)}{=} \frac{2}{\sqrt{3}} \arctan{ \left( \frac{2y+1}{\sqrt{3}} \right) } - \int \frac{t - \frac{1}{\sqrt{3}}}{t^2+1} \mathrm{d}t \\
&=\frac{2}{\sqrt{3}} \arctan{ \left( \frac{2y+1}{\sqrt{3}} \right) } - \frac{1}{2}\log({t^2+1}) + \frac{1}{\sqrt{3}}\arctan{t} \\
&=\sqrt{3}\arctan{ \left( \frac{2y+1}{\sqrt{3}} \right) } - \frac{1}{2}\log(y^2+y+1)
\end{align*}

\( \displaystyle *: \) In the second integral we use the substitution \( \displaystyle u = \frac{2t+1}{\sqrt{3}} \)

Hence,

\[ \displaystyle
\frac{1}{2}\int \frac{1-y}{y^2+y+1 } \mathrm{d}y + \frac{1}{2}\int\frac{\mathrm{d}y}{y-1} = \sqrt{3}\arctan{ \left( \frac{2y+1}{\sqrt{3}} \right) } - \frac{1}{2}\log(y^2+y+1) +\frac{1}{2}\log|y-1|
\]

From now on, it's just a matter of calculations. Taking into account that \( \displaystyle y = \sqrt[3]{u} \text{ and } u = x^2 + 1 \) we can compute the given integral in terms of \( \displaystyle x. \)


Thanks again for the solutions!
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Tolaso J Kos
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Re: Some indefinite integrals

#5

Post by Tolaso J Kos »

... The time is so past ....
One attempt to evaluate the given integral using the sub
$x=\tan u$.

$$\begin{aligned}
\int \frac{dx}{x\sqrt[3]{x^2+1}} &\overset{x=\tan u}{=\! =\! =\! =\!}\int \frac{\sec^2 u}{\tan u\sqrt[3]{\tan^2 u+1}}\, du \\
&\overset{\tan^2 u+1 =\sec^2 u}{=\! =\! =\! =\! =\! =\! =\! =\! =\!}\int \frac{\sec^2 u}{\tan x\sqrt[3]{\sec^2 u}}\, du \\
&= \int \frac{\sqrt[3]{\sec^4 u}}{\tan u}\, du\\
&= \int \sqrt[3]{\frac{\sec^4 u}{\tan^3 u}}\, du\\
&= \int \sqrt[3]{\frac{1}{\cos^4 u}\cdot \frac{\cos^3 u}{\sin^3 u}}\, du\\
&= \int \sqrt[3]{\frac{\sec u}{\sin^3 u}}\, du\\
&= \int \frac{\sqrt[3]{\sec u}}{\sin u}\, du
\end{aligned}$$

The last integral remaining is indeed a toughie. Now one way to go around it would be by multiplying nominator and denominator by $-\sin^2 u\sqrt[3]{\cos u}$. This will give us:

$$\int \frac{-\sin u}{-\sqrt[3]{\cos u}\sin^2 u }\, du=\int \frac{-\sin u}{\sqrt[3]{\cos u}\left ( \cos^2 u-1 \right )}\, du\overset{y=\cos u}{=\! =\! =\! =\!}\int \frac{dy}{\sqrt[3]{y}\left ( y^2-1 \right )}\overset{t=\sqrt[3]{y}}{=\! =\! =\! }3\int \frac{t}{t^6-1}\, dt=\cdots$$

For the last integral we will use partial decomposition...
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Papapetros Vaggelis
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Re: Some indefinite integrals

#6

Post by Papapetros Vaggelis »

Solution for ii)

Apostolos, we have \(\displaystyle{2+\sin\,(2\,x)}\).

The integration interval is \(\displaystyle{I=\mathbb{R}}\) .

\(\displaystyle{2+\sin\,(2\,x)=1+\left(\sin\,x+\cos\,x\right)^2=1+2\,\sin^2\,\left(x+\frac{\pi}{4}\right)\,,x\in\mathbb{R}}\) , so

\(\displaystyle{\int \dfrac{\sin\,x}{\sqrt{2+\sin\,(2\,x)}}\,\rm{dx}=\int \dfrac{\sin\,x}{\sqrt{1+2\,\sin^2\,\left(x+\frac{\pi}{4}\right)}}\,\rm{dx}}\) .

By substituting \(\displaystyle{t=x+\frac{\pi}{4}}\), we have that \(\displaystyle{\rm{dt}=\rm{dx}}\) and \(\displaystyle{\sin\,x=\dfrac{\sin\,t-\cos\,t}{\sqrt{2}}}\) , so

\(\displaystyle{\begin{aligned}\int \frac{\sin\,x}{\sqrt{2+\sin\,(2\,x)}}\,\rm{dx}&=\frac{1}{\sqrt{2}}\,\int \frac{\sin\,t-\cos\,t}{\sqrt{1+2\,\sin^2\,t}}\,\rm{dt}\\&=\frac{1}{\sqrt{2}}\,\left(\int \frac{\sin\,t}{\sqrt{3-2\,\cos^2\,t}}\,\rm{dt}-\int \frac{\cos\,t}{\sqrt{1+\left(\sqrt{2}\,\sin\,t\right)^2}}\,\rm{dt}\right)\\&=\frac{1}{2}\,\left(\int \frac{\sqrt{2}\,\sin\,t}{\sqrt{3}\,\sqrt{1-\left(\frac{\sqrt{2}\,\cos\,t}{\sqrt{3}}\right)^2}}\,\rm{dt}-\int \frac{\sqrt{2}\,\cos\,t}{\sqrt{1+\left(\sqrt{2}\,\sin\,t\right)^2}}\,\rm{dt}\right)\\&=\frac{1}{2}\,\left(\arccos\,\left(\dfrac{\sqrt{2}\,\cos\,t}{\sqrt{3}}\right)-{\rm{arcsinh}}\left(\sqrt{2}\,\sin\,t\right)\right)+c\\&=\frac{1}{2}\,\left(\arccos\,\left(\dfrac{\cos\,x-\sin\,x}{\sqrt{3}}\right)-{\rm{arcsinh}}\left(\sin\,x+\cos\,x\right)\right)+c\end{aligned}}\)
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Re: Some indefinite integrals

#7

Post by Tsakanikas Nickos »

Thank you, Mr.Papapetros, firstly for your nice solution and secondly for giving the exact solution of the problem (ii).
I didn't notice the "mistake" that Tolaso's solution contained, which you pointed out!

I'm glad that we had such a long and interesting discussion and slightly different points of view on the second and third integral!
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