Some Indefinite Integrals

Calculus (Integrals, Series)
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jacks
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Some Indefinite Integrals

#1

Post by jacks »

Evaluation of Following Indefinite Integrals:

(a) \(\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx\)

(b) \(\displaystyle \int\frac{1+x^{-\frac{2}{3}}}{1+x}dx\)

(c) \(\displaystyle \int \frac{x\cdot \ln(x)}{(x^2-1)^{\frac{3}{2}}}\)

(d) \(\displaystyle \int\frac{x^2+20}{\left(x\cdot \sin x+5\cdot \cos x\right)^2}dx\)
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Tolaso J Kos
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Re: Some Indefinite Integrals

#2

Post by Tolaso J Kos »

(c) It seems so familiar, yet I do not remember where I've seen it before. Anyway.
Note that: \( \displaystyle \left ( -\frac{1}{\sqrt{x^2-1}} \right )'=\frac{x}{\left ( x^2-1 \right )^{3/2}} \).

Apply IBP to get:
$$\begin{aligned}
\int \frac{x\ln x}{\left ( x^2-1 \right )^{3/2}}\, dx &=-\frac{\ln x}{\sqrt{x^2-1}}+\int \frac{dx}{x\sqrt{x^2-1}} \\
&\overset{u=\sqrt{x^2-1}}{=\! =\! =\! =\! =\! =\!}-\frac{\ln x}{\sqrt{x^2-1}}+\int \frac{du}{u^2+1} \\
&= -\frac{\ln x}{\sqrt{x^2-1}}+\tan^{-1}u\\
&= -\frac{\ln x}{\sqrt{x^2-1}}+\tan^{-1}\sqrt{x^2-1}+c
\end{aligned}$$

(b) Applying partial fractions you get: \( \displaystyle \frac{1+x^{-2/3}}{x+1}=\frac{1}{x+1}+\frac{1}{x^{2/3}(x+1)} \).
Integrating the last equation you get:
$$\int \frac{1+x^{-2/3}}{x+1}\, dx=\int \frac{1}{x+1}\, dx+\int \frac{1}{x^{2/3}\, dx(x+1)}$$
$$\int \frac{1+x^{-2/3}}{x+1}\, dx=\ln \left | x+1 \right |+\int \frac{1}{x^{2/3}\, dx(x+1)}$$
$$\begin{aligned}
\int \frac{dx}{x^{2/3}(x+1)}\, dx &\overset{u=\sqrt[3]{x}}{=\! =\! =\! =\!}3\int \frac{du}{u^3+1}=3\int \left ( \frac{2-u}{3\left ( u^2-u+1 \right )}+\frac{1}{3\left ( u+1\right )} \right )\, du \\
&=\int \frac{2-u}{u^2-u+1}\, du+\int \frac{du}{u+1} \\
&= \int \left ( \frac{3}{2\left ( u^2-u+1 \right )}-\frac{2u-1}{2\left ( u^2-u+1 \right )} \right )\, du+\ln \left | u+1 \right |\\
&= \frac{3}{2}\int \frac{du}{u^2-u+1}-\frac{1}{2}\int \frac{2u-1}{u^2-u+1}\, du+\ln \left | u+1 \right |\\
&=\frac{3}{2}\int \frac{du}{\left ( u-\frac{1}{2} \right )^2+\frac{3}{4}}-\frac{1}{2}\ln \left | u^2-u+1 \right |+\ln \left | u+1 \right |
\end{aligned}$$

I think you can take it from here. Don't forget to substitute \(x \) back again to what you'll find...
Imagination is much more important than knowledge.
jacks
Posts: 102
Joined: Thu Nov 12, 2015 5:26 pm
Location: Himachal Pradesh (INDIA)

Re: Some Indefinite Integrals

#3

Post by jacks »

**My Solution ** For \(\displaystyle (d)\;\;\ I = \int\frac{x^2+20}{\left(x\cdot \sin x+5\cdot \cos x\right)^2}dx\)

Now \(\displaystyle \left(x\cdot \sin x+5\cdot \cos x\right) = \sqrt{x^2+25}\left(\frac{x}{\sqrt{x^2+25}}\cdot \sin x + \frac{5}{x^2+25}\cdot \cos x\right) = \sqrt{x^2+25}\cdot \cos \left(x-\phi\right)\)

Where \(\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+25}}\) and \(\displaystyle \cos \phi = \frac{5}{\sqrt{x^2+25}}\) and \(\displaystyle \tan \phi = \frac{x}{5}\)

So Integral \(\displaystyle I = \int \sec^2 \left(x-\phi\right)\cdot \left(\frac{x^2+20}{x^2+25}\right)dx\)

Now Let \(\displaystyle \left(x-\phi\right) = \left(x-\tan^{-1}\left(\frac{x}{5}\right)\right) = t\), Then \(\displaystyle dx = \left(\frac{x^2+20}{x^2+25}\right)\)

So \(\displaystyle I = \int \sec^2 t dt = \tan t +\mathbb{C} = \tan \left(x-\tan^{-1}\left(\frac{x}{5}\right)\right)+\mathbb{C} = \left(\frac{5\tan x-x}{x\tan x+5}\right)+\mathbb{C}\)

So \(\displaystyle I =\int\frac{x^2+20}{\left(x\cdot \sin x+5\cdot \cos x\right)^2}dx = \left(\frac{5\tan x-x}{x\tan x+5}\right)+\mathbb{C}\)
jacks
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Re: Some Indefinite Integrals

#4

Post by jacks »

**My Solution for \(\displaystyle (a):: I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \frac{\cos 2x}{\sin x\cdot \sqrt{\cos 2x}}\cdot \frac{\sin x}{\sin x}dx = \int\frac{\sqrt{2\cos^2 x-1}}{\left(1-\cos^2 x\right)\cdot \sqrt{2\cos^2 x-1}}\cdot \sin x dx\)

Now Let \(\displaystyle \cos x = t\) and \(\displaystyle \sin xdx = -dt\)

So Integral \(\displaystyle I = \int\frac{\left(2t^2-1\right)}{(t^2-1)\cdot \sqrt{2t^2-1}}dt\)

So \(\displaystyle I = \int \frac{2(t^2-1)+1}{(t^2-1)\sqrt{2t^2-1}}dt = \sqrt{2}\int \frac{1}{\sqrt{t^2-\left(\frac{1}{\sqrt{2}}\right)^2}}dt +\int \frac{1}{(t^2-1)\sqrt{2t^2-1}}dt\)

\(\displaystyle I = \sqrt{2}\cdot \left|t+\sqrt{t^2-\left(\frac{1}{\sqrt{2}}\right)^2}\right|+J\)


Now for calculation of \(\displaystyle J = \int\frac{1}{(t^2-1)\sqrt{2t^2-1}}dt\), Put \(\displaystyle t=\frac{1}{u}\) Then \(\displaystyle dt = -\frac{1}{u^2}du\)

So \(\displaystyle J = -\int\frac{u}{(1-u^2)\sqrt{2-u^2}}du\), Now again put \((2-u^2) = v^2\) Then \(\displaystyle udu = -vdv\)

So \(\displaystyle J = \int\frac{1}{v^2-1}dv = \frac{1}{2}\ln \left|\frac{v-1}{v+1}\right| = \frac{1}{2}\ln \left|\frac{\sqrt{2-u^2}-1}{\sqrt{2+u^2}+1}\right|\)

So \(\displaystyle J = \frac{1}{2}\ln \left|\frac{\sqrt{2t^2-1}-1}{\sqrt{2t^2+1}+1}\right| = \frac{1}{2}\ln \left|\frac{\sqrt{2\cos^2 x-1}-1}{\sqrt{2\cos^2 x+1}+1}\right|\)

So \(\displaystyle I = \sqrt{2}\ln \left|\frac{2\cos x}{2} +\frac{\sqrt{2\cos^2 x - 1}}{2}\right|+\frac{1}{2}\ln \left|\frac{\sqrt{2\cos^2 x-1}-1}{\sqrt{2\cos^2 x+1}+1}\right|+\mathbb{C}\)

So \(\displaystyle I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \sqrt{2}\ln \left|2\cos x+\sqrt{\cos 2x}\right|+\frac{1}{2}\ln \left|\frac{\sqrt{\cos 2x}-1}{\sqrt{\cos 2x}+1}\right|+\mathbb{C}\)

But above method is very Lengthy, If anyone have a better solution , Then please write here

Thanks
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