Some indefinite integrals
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Some indefinite integrals
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) \( \displaystyle \int \frac{1}{x\sqrt[3]{x^{2}+1}} \mathrm{d}x \)
(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) \( \displaystyle \int \frac{1}{x\sqrt[3]{x^{2}+1}} \mathrm{d}x \)
- Grigorios Kostakos
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Re: Some indefinite integrals
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*}
\(i.\) \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*}
Grigorios Kostakos
- Tolaso J Kos
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Re: Some indefinite integrals
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.
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.
Imagination is much more important than knowledge.
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Re: Some indefinite integrals
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!
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!
- Tolaso J Kos
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Re: Some indefinite integrals
... 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...
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...
Imagination is much more important than knowledge.
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Re: Some indefinite integrals
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
\(\displaystyle{\begin{aligned}\int \frac{\sin\,x}{\sqrt{2+\sin\,(2\,x)}}\,\rm{dx}&=\frac{1}{\sqrt{2}}\,\int \frac{\sin\,t-\cos\,t}{\sqrt{1+2\,\sin^2\,t}}\,\rm{dt}\\&=\frac{1}{\sqrt{2}}\,\left(\int \frac{\sin\,t}{\sqrt{3-2\,\cos^2\,t}}\,\rm{dt}-\int \frac{\cos\,t}{\sqrt{1+\left(\sqrt{2}\,\sin\,t\right)^2}}\,\rm{dt}\right)\\&=\frac{1}{2}\,\left(\int \frac{\sqrt{2}\,\sin\,t}{\sqrt{3}\,\sqrt{1-\left(\frac{\sqrt{2}\,\cos\,t}{\sqrt{3}}\right)^2}}\,\rm{dt}-\int \frac{\sqrt{2}\,\cos\,t}{\sqrt{1+\left(\sqrt{2}\,\sin\,t\right)^2}}\,\rm{dt}\right)\\&=\frac{1}{2}\,\left(\arccos\,\left(\dfrac{\sqrt{2}\,\cos\,t}{\sqrt{3}}\right)-{\rm{arcsinh}}\left(\sqrt{2}\,\sin\,t\right)\right)+c\\&=\frac{1}{2}\,\left(\arccos\,\left(\dfrac{\cos\,x-\sin\,x}{\sqrt{3}}\right)-{\rm{arcsinh}}\left(\sin\,x+\cos\,x\right)\right)+c\end{aligned}}\)
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
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!
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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