Trigonometric Indefinite Integral

Real Analysis
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jacks
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Trigonometric Indefinite Integral

#1

Post by jacks »

Calculation of \(\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\)

**My Try*: Let \(\displaystyle I = \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int\frac{\sqrt{\tan x}}{\sqrt{\tan x}+1}dx = x+\int\frac{1}{1+\sqrt{\tan x}}dx\)

Now Let \(\displaystyle \tan x= t^2\;,\) Then \(\displaystyle sec^2 xdx = 2tdt\Rightarrow dx = \frac{2t}{1+t^4}dt\)

So \(\displaystyle I = x+2\int\frac{t}{(1+t^4)(1+t)}dt = x+2\int\frac{(1+t)-1}{(1+t^4)(1+t)}dt = x+2\int\frac{1}{1+t^4}dt-2\int\frac{1}{(1+t)(1+t^4)}dt\)

Now I did not Understand How can I calculate \(\displaystyle \int \frac{1}{(1+t)(1+t^4)}dt\)

Help me

Thanks
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Tolaso J Kos
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Re: Trigonometric Indefinite Integral

#2

Post by Tolaso J Kos »

Well Jacks, you almost got it.
For the last integral use partial decomposition.
Note that:
$$\frac{1}{\left ( 1+x \right )\left ( 1+x^4 \right )}=\frac{1-x}{4\left ( x^2+\sqrt{2}x+1 \right )}+\frac{x-1}{4\left ( -x^2+\sqrt{2}x-1 \right )}+\frac{1}{2\left ( x+1 \right )}$$

Now integrate term by term.
Note (again) that:
$$\frac{1-x}{x^2+\sqrt{2}x+1}=\frac{-2x+\sqrt{2}}{2\left ( x^2+\sqrt{2}x+1 \right )}-\frac{\sqrt{2}+2}{2\left ( x^2+\sqrt{2}x+1 \right )}$$

I think I covered you. Now everything else that will occur is easy.

Note It is very interesting to note that the definite integral, \( \displaystyle \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\, dx \) can be handled in much easier way. The trick is to apply the sub \( u=\pi/2-x \) and get the final result: \( \pi/4 \). This is left as an exercise.
Imagination is much more important than knowledge.
jacks
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Location: Himachal Pradesh (INDIA)

Re: Trigonometric Indefinite Integral

#3

Post by jacks »

Thanks Tolaso J. Kos
Papapetros Vaggelis
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Re: Trigonometric Indefinite Integral

#4

Post by Papapetros Vaggelis »

Tolaso J Kos wrote:Note It is very interesting to note that the definite integral, \( \displaystyle \int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\, dx \) can be handled in much easier way. The trick is to apply the sub \( u=\pi/2-x \) and get the final result: \( \pi/4 \). This is left as an exercise.
Ok, let's prove that

\(\displaystyle{\int_{0}^{\frac{\pi}{2}}\dfrac{\sqrt{\sin\,x}}{\sqrt{\cos\,x}+\sqrt{\sin\,x}}\,\mathrm{d}x=\frac{\pi}{4}}\) .

As Tolis said, we apply the sub \(\displaystyle{x=\frac{\pi}{2}-u}\) and we get :

\(\displaystyle{\begin{aligned} J&=-\int_{\frac{\pi}{2}}^{0}\dfrac{\sqrt{\sin\,\left(\frac{\pi}{2}-u\right)}}{\sqrt{\cos\,\left(\frac{\pi}{2}-u\right)}+\sqrt{\sin\,\left(\frac{\pi}{2}-u\right)}}\,\mathrm{d}u\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{\sqrt{\cos\,u}}{\sqrt{\cos\,u}+\sqrt{\sin\,u}}\,\mathrm{d}u\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{\sqrt{\cos\,x}}{\sqrt{\cos\,x}+\sqrt{\sin\,x}}\,\mathrm{d}x\end{aligned}}\)

so,

\(\displaystyle{2\,J=\int_{0}^{\frac{\pi}{2}}\dfrac{\sqrt{\sin\,x}+\sqrt{\cos\,x}}{\sqrt{\cos\,x}+\sqrt{\sin\,x}}\,\mathrm{d}x=\frac{\pi}{2}\implies J=\frac{\pi}{4}}\)
Papapetros Vaggelis
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Re: Trigonometric Indefinite Integral

#5

Post by Papapetros Vaggelis »

Hello jacks.

Firstly, \(\displaystyle{I=x-\int \dfrac{1}{\sqrt{\tan\,x}+1}\,\mathrm{d}x}\) .

By substituting \(\displaystyle{u=\sqrt{\tan\,x}}\) , we get \(\displaystyle{\tan\,x=u^2}\) and then

\(\displaystyle{x=\arctan\,u^2\implies \mathrm{d}x=\dfrac{2\,u}{1+u^4}\,\mathrm{d}u}\) , so

\(\displaystyle{\int \dfrac{1}{1+\sqrt{\tan\,x}}\,\mathrm{d}x=\int \dfrac{2\,u}{\left(1+u\right)\,\left(u^4+1\right)}\,\mathrm{d}u}\) .

For each \(\displaystyle{u\in\mathbb{R}}\) holds : \(\displaystyle{u^4+1=\left(u^2+\sqrt{2}\,u+1\right)\,\left(u^2-\sqrt{2}\,u+1\right)}\) , so

\(\displaystyle{\dfrac{2\,u}{\left(1+u\right)\,\left(1+u^4\right)}=\dfrac{a}{1+u}+\dfrac{b\,u+c}{u^2+\sqrt{2}\,u+1}+\dfrac{d\,u+e}{u^2-\sqrt{2}\,u+1}}\)

for some \(\displaystyle{a\,,b\,,c\,,d\,,e\in\mathbb{R}}\) .
jacks
Posts: 102
Joined: Thu Nov 12, 2015 5:26 pm
Location: Himachal Pradesh (INDIA)

Re: Trigonometric Indefinite Integral

#6

Post by jacks »

Thanks Papapetros Vaggelis
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