Some Definite Integrals

Calculus (Integrals, Series)
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jacks
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Some Definite Integrals

#1

Post by jacks »

(1) \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^9 x}{\sin^3 x+\cos^3 x}dx\)

(2) \(\displaystyle \int_{0}^{a}x^3\cdot \sqrt{2ax-x^2}dx\)

(3) \(\displaystyle \int_{0}^{1}\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx\)

(4) \(\displaystyle\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\left(\frac{x^4}{1-x^4}\right)\cdot \cos^{-1}\left(\frac{2x}{1+x^2}\right)dx \)

(5) \(\displaystyle \int_{0}^{1}\frac{x^4\cdot \left(1-x\right)^4}{1+x^2}dx\)
Papapetros Vaggelis
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Re: Some Definite Integrals

#2

Post by Papapetros Vaggelis »

\(\displaystyle{\bullet\,\,(1)}\)

The integrand function is well defined and continuous at \(\displaystyle{\left[0,\frac{\pi}{2}\right]}\) , so

\(\displaystyle{I_{1}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^9\,x}{\sin^3\,x+\cos^3\,x}\,\mathrm{d}x<\infty}\) .

We apply the sub \(\displaystyle{t=\frac{\pi}{2}-x\,,t\in\left[0,\frac{\pi}{2}\right]}\) and we get :

\(\displaystyle{\begin{aligned}I_{1}&=-\int_{\frac{\pi}{2}}^{0}\dfrac{\sin^9\,\left(\frac{\pi}{2}-t\right)}{\sin^3\,\left(\frac{\pi}{2}-t\right)+\cos^3\,\left(\frac{\pi}{2}-t\right)}\,\mathrm{d}t\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^9\,t}{\sin^3\,t+\cos^3\,t}\,\mathrm{d}t\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^9\,x}{\sin^3\,x+\cos^3\,x}\,\mathrm{d}x\end{aligned}}\)

By adding :

\(\displaystyle{\begin{aligned} 2\,I_{1}&=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^9\,x+\cos^9\,x}{\sin^3\,x+\cos^3\,x}\,\mathrm{d}x\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{\left(\sin^3\,x+\cos^3\,x\right)\,\left(\sin^6\,x-\sin^3\,x\,\cos^3\,x+\cos^6\,x\right)}{\sin^3\,x+\cos^3\,x}\,\mathrm{d}x\\&=\int_{0}^{\frac{\pi}{2}}\left[\left(\sin^2\,x+\cos^2\,x\right)\,\left(\sin^4\,x-\sin^2\,x\,\cos^2\,x+\cos^4\,x\right)-\sin^3\,x\,\cos^3\,x\right]\,\mathrm{d}x\\&=\int_{0}^{\frac{\pi}{2}}\left[\left(\sin^2\,x+\cos^2\,x\right)^2-3\,\sin^2\,x\,\cos^2\,x-\sin^3\,x\,\cos^3\,x\right]\,\mathrm{d}x\\&=\int_{0}^{\frac{\pi}{2}}\left(1-3\,\sin^2\,x\,\cos^2\,x-\sin^3\,x\,\cos^3\,x\right)\,\mathrm{d}x\\&=\dfrac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}3\,\sin^2\,x\,\cos^2\,x\,\mathrm{d}x-\int_{0}^{\frac{\pi}{2}}\sin^3\,x\,\cos^3\,x\,\mathrm{d}x\\&=\dfrac{\pi}{2}+\left[\sin\,x\,\cos^3\,x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\cos^4\,x\,\mathrm{d}x+\left[\dfrac{1}{4}\,\sin^2\,x\,\cos^4\,x\right]_{0}^{\frac{\pi}{2}}-\dfrac{1}{2}\,\int_{0}^{\frac{\pi}{2}}\sin\,x\,\cos^5\,x\,\mathrm{d}x\\&=\dfrac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}\left(\dfrac{1+\cos\,2\,x}{2}\right)^2\,\mathrm{d}x+\left[\dfrac{\cos^6\,x}{12}\right]_{0}^{\frac{\pi}{2}}\\&=\dfrac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}\left(\dfrac{1}{4}+\dfrac{\cos\,2\,x}{2}+\dfrac{\cos^2\,2\,x}{4}\right)\,\mathrm{d}x-\dfrac{1}{12}\\&=\dfrac{\pi}{2}-\int_{0}^{\frac{\pi}{2}}\left(\dfrac{1}{4}+\dfrac{\cos\,2\,x}{2}+\dfrac{1+\cos\,4\,x}{8}\right)\,\mathrm{d}x-\dfrac{1}{12}\\&=\dfrac{\pi}{2}-\left[\dfrac{3\,x}{8}+\dfrac{\sin\,2\,x}{4}+\dfrac{\sin\,4\,x}{32}\right]_{0}^{\frac{\pi}{2}}-\dfrac{1}{12}\\&=\dfrac{\pi}{2}-\dfrac{3\,\pi}{16}-\dfrac{1}{12}\\&=\dfrac{24\,\pi-9\,\pi-4}{48}\\&=\dfrac{15\,\pi-4}{48}\end{aligned}}\)

so \(\displaystyle{I_{1}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^9\,x}{\sin^3\,x+\cos^3\,x}\,\mathrm{d}x=\dfrac{15\,\pi-4}{96}}\)



\(\displaystyle{\bullet\,\,(2)}\)

The integrand function is well defined and continuous at \(\displaystyle{\left[0,a\right]}\) , so

\(\displaystyle{I_{2}=\int_{0}^{a}x^3\,\sqrt{a^2-\left(a-x\right)^2}\,\mathrm{d}x<\infty}\) .

By substituting \(\displaystyle{a-x=a\,\sin\,t}\) , we have :

\(\displaystyle{\begin{aligned} I_{2}&=-a\,\int_{\frac{\pi}{2}}^{0}\left(a-a\,\sin\,t\right)^3\,\sqrt{a^2-a^2\,\sin^2\,t}\,\cos\,t\,\mathrm{d}t\\&=a^5\,\int_{0}^{\frac{\pi}{2}}\left(1-\sin\,t\right)^3\,\cos^2\,t\,\mathrm{d}t\\&=a^5\,\int_{0}^{\frac{\pi}{2}}\,\left[\cos^2\,t-3\,\sin\,t\,\cos^2\,t+3\,\sin^2\,t\,\cos^2\,t-\sin^3\,t\,\cos^2\,t\right]\,\mathrm{d}t\\&=a^5\,\int_{0}^{\frac{\pi}{2}}\dfrac{1+\cos\,2\,t}{2}\,\mathrm{d}t+\left[a^5\,\cos^3\,t\right]_{0}^{\frac{\pi}{2}}+a^5\,\left[\left[\cos\,t\,\sin^3\,t\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sin^4\,t\,\mathrm{d}t\right]+\\
&\qquad a^5\,\left[\left[\dfrac{1}{3}\,\sin^2\,t\,\cos^3\,t\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\,\dfrac{2}{3}\sin\,t\,\cos^4\,t\,\mathrm{d}t\right]\\&=a^5\,\left[\dfrac{t}{2}+\dfrac{\sin\,2\,t}{4}\right]_{0}^{\frac{\pi}{2}}-a^5+a^5\,\int_{0}^{\frac{\pi}{2}}\,\left(\dfrac{1-\cos\,2\,t}{2}\right)^2\,\mathrm{d}t+\left[\dfrac{2\,a^5}{15}\,\cos^5\,t\right]_{0}^{\frac{\pi}{2}}\\&=\dfrac{\pi\,a^5}{4}-a^5+a^5\,\int_{0}^{\frac{\pi}{2}}\left(\dfrac{1}{4}-\dfrac{\cos\,2\,t}{2}+\dfrac{\cos^2\,2\,t}{4}\right)\,\mathrm{d}t-\dfrac{2\,a^5}{15}\\&=\dfrac{\pi\,a^5}{4}-\dfrac{17\,a^5}{15}+a^5\,\left[\dfrac{3\,t}{8}+\dfrac{\sin\,4\,t}{32}-\dfrac{\sin\,2\,t}{4}\right]_{0}^{\frac{\pi}{2}}\\&=\dfrac{\pi\,a^5}{4}-\dfrac{17\,a^5}{15}+\dfrac{3\,\pi\,a^5}{16}\\&=\dfrac{a^5}{240}\,\left(105\,\pi-272\right)\end{aligned}}\)

Similar problem definite integral formula
Papapetros Vaggelis
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Re: Some Definite Integrals

#3

Post by Papapetros Vaggelis »

\(\displaystyle{\bullet\,\,(5)}\)

\(\displaystyle{\left(1-x\right)^4=\left(x-1\right)^4=\left[\left(x^2+1\right)-2\,x\right]^2=\left(x^2+1\right)^2-4\,x\,\left(x^2+1\right)+4\,x^2}\) , so :

\(\displaystyle{\begin{aligned}I_{5}&=\int_{0}^{1}\dfrac{x^4\,\left(x^2+1\right)^2-4\,x^5\,\left(x^2+1\right)+4\,x^6}{x^2+1}\,\mathrm{d}x\\&=\int_{0}^{1}\left(x^4\,\left(x^2+1\right)-4\,x^5+4\,\dfrac{\left(x^4-x^2+1\right)\,\left(x^2+1\right)-1}{x^2+1}\right)\,\mathrm{d}x\\&=\int_{0}^{1}\left(x^6+x^4-4\,x^5+4\,x^4-4\,x^2+4-\dfrac{4}{x^2+1}\right)\,\mathrm{d}x\\&=\left[\dfrac{x^7}{7}-\dfrac{4\,x^6}{6}+x^5-\dfrac{4\,x^3}{3}+4\,x-4\,\arctan\,x\right]_{0}^{1}\\&=\dfrac{1}{7}-\dfrac{2}{3}+1-\dfrac{4}{3}+4-\pi\\&=\dfrac{22}{7}-\pi\end{aligned}}\)
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Re: Some Definite Integrals

#4

Post by Papapetros Vaggelis »

\(\displaystyle{\bullet\,\,(3)}\)

The function \(\displaystyle{I:\left[0,1\right)\longrightarrow \mathbb{R}\,\,,x\mapsto \int_{0}^{x}\dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}\,\mathrm{d}t}\)

is well defined because for each \(\displaystyle{x\in\left[0,1\right)}\) the function \(\displaystyle{g:\left[0,x\right]\longrightarrow \mathbb{R}\,\,,t\mapsto \dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}}\)

is continuous at \(\displaystyle{\left[0,x\right]}\) and thus \(\displaystyle{I(x)=\int_{0}^{x}\dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}\,\mathrm{d}t\in\mathbb{R}}\) .

If \(\displaystyle{\lim_{x\to 1^{-}}I(x)\in\mathbb{R}}\) , then \(\displaystyle{\int_{0}^{1}\dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}\,\mathrm{d}t\in\mathbb{R}}\)

and \(\displaystyle{\lim_{x\to 1^{-}}I(x)=\int_{0}^{1}\dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}\,\mathrm{d}t}\) .

In order to calculate the integral \(\displaystyle{I(x)=\int_{0}^{x}\dfrac{2-t^2}{\left(1+t\right)\,\sqrt{1-t^2}}\,\mathrm{d}t\,,x\in\left[0,1\right)}\) ,

we apply the substitution \(\displaystyle{y=\arcsin\,t\,,t\in\left[0,\arcsin\,x\right]}\)

and we get : \(\displaystyle{\mathrm{d}y=\dfrac{1}{\sqrt{1-t^2}}\,\mathrm{d}t\,\,,t=\sin,y}\) , so :

\(\displaystyle{\begin{aligned} I(x)&=\int_{0}^{\arcsin\,x} \dfrac{2-\sin^2\,y}{1+\sin\,y}\,\mathrm{d}y\\&=\int_{0}^{\arcsin\,x}\left(\dfrac{1}{1+\sin\,y}+\dfrac{1-\sin^2\,y}{1+\sin\,y}\right)\,\mathrm{d}y\\&=\int_{0}^{\arcsin\,x}\left(\dfrac{1}{\left(\sin\,\dfrac{y}{2}+\cos\,\dfrac{y}{2}\right)^2}+\dfrac{\left(1-\sin\,y\right)\,\left(1+\sin\,y\right)}{1+\sin\,y}\right)\,\mathrm{d}y\\&=\int_{0}^{\arcsin\,x}\left(\dfrac{1}{2\,\cos^2\,\left(\dfrac{y}{2}-\dfrac{\pi}{4}\right)}+1-\sin\,y\right)\,\mathrm{d}y\\&=\left[\tan\,\left(\dfrac{y}{2}-\dfrac{\pi}{4}\right)+y+\cos\,y\right]_{0}^{\arcsin\,x}\\&=\tan\,\left(\dfrac{\arcsin\,x}{2}-\dfrac{\pi}{4}\right)+\arcsin\,x+\cos\,\left(\arcsin\,x\right)+1-0-\cos\,0\\&=\tan\,\left(\dfrac{\arcsin\,x}{2}-\dfrac{\pi}{4}\right)+\arcsin\,x+\cos\,\left(\arcsin\,x\right)\end{aligned}}\)

Thus,

\(\displaystyle{\begin{aligned} \lim_{x\to 1^{-}}I(x)&=\lim_{x\to 1^{-}}\left(\tan\,\left(\dfrac{\arcsin\,x}{2}-\dfrac{\pi}{4}\right)+\arcsin\,x+\cos\,\left(\arcsin\,x\right)\right)\\&=\tan\,\left(\dfrac{\arcsin\,1}{2}-\dfrac{\pi}{4}\right)+\arcsin\,1+\cos\,\left(\arcsin\,1\right)\\&=\tan\,0+\dfrac{\pi}{2}+\cos\,\dfrac{\pi}{2}\\&=\dfrac{\pi}{2}\end{aligned}}\)

Result : \(\displaystyle{\int_{0}^{1}\dfrac{2-x^2}{\left(1+x\right)\,\sqrt{1-x^2}}\,\mathrm{d}x=\lim_{x\to 1^{-}}I(x)=\dfrac{\pi}{2}}\) .
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Tolaso J Kos
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Re: Some Definite Integrals

#5

Post by Tolaso J Kos »

(4):
This is kinda fun. Here is an approach.

$$\begin{aligned}
I=\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\cos^{-1}\left ( \frac{2x}{1+x^2} \right )\,dx &= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\cos^{-1}\left ( -\frac{2x}{1+x^4} \right )\,dx\\
&= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\left ( \pi-\cos^{-1}\left ( \frac{2x}{1+x^4} \right ) \right )\,dx\\
&=\pi\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,dx-I
\end{aligned}$$

Thus:
$$I=\frac{\pi}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,dx\implies I=-\frac{2}{\sqrt{3}}+\frac{\pi}{6}+\frac{1}{2}\ln \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )$$
:)

At first I applied parts, which is not such a nice solution! The good news is that the derivative of \( \cos^{-1} \) is good, but ...
Imagination is much more important than knowledge.
Papapetros Vaggelis
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Re: Some Definite Integrals

#6

Post by Papapetros Vaggelis »

2nd solution for \(\displaystyle{I=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)\,\mathrm{d}x}\) .

\(\displaystyle{\bullet\,\,(4) : I=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)\,\mathrm{d}x}\)

At first, I tried to find the derivative of \(\displaystyle{\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)}\) which is

\(\displaystyle{-\dfrac{2}{1+x^2}\,,x\in\left[-\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right]}\) .

So, there is \(\displaystyle{c\in\mathbb{R}}\) such that

\(\displaystyle{\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)=c-2\,\arctan\,x\,\forall\,x\in\left[-\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right]}\) .

If \(\displaystyle{x=0}\) , then \(\displaystyle{\arccos\,0=c-2\,\arctan\,0\iff c=\pi}\) , so :

\(\displaystyle{\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)=\pi-2\,\arctan\,x\,,x\in\left[-\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right]}\) .

Note : \(\displaystyle{-\dfrac{1}{\sqrt{3}}\leq x\leq \dfrac{1}{\sqrt{3}}\implies \dfrac{2\,x}{1+x^2}\in\left[-\dfrac{\sqrt{3}}{2},\dfrac{\sqrt{3}}{2}\right]\subset \left[-1,1\right]}\) .

Now, \(\displaystyle{I=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\left(\pi-2\,\arctan\,x\right)\,\mathrm{d}x}\) .

We are using the identity \(\displaystyle{\arctan\,(-x)=-\arctan\,x\,,x\in\mathbb{R}}\) .

By substituting \(\displaystyle{x=-u}\) we get :

\(\displaystyle{I=\int_{\frac{1}{\sqrt{3}}}^{-\frac{1}{\sqrt{3}}}\dfrac{(-u)^4}{1-(-u)^4}\,\left(\pi-2\,\arctan\,(-u)\right)\,(-\mathrm{d}u)=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{u^4}{1-u^4}\left(\pi+2\,\arctan\,u\right)\,\mathrm{d}u}\)

Therefore,

\(\displaystyle{\begin{aligned} 2\,I&=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\left(\pi-2\,\arctan\,x+\pi+2\,\arctan\,x\right)\,\mathrm{d}x\\&=2\,\pi\,\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\mathrm{d}x\\&=4\,\pi\,\int_{0}^{\frac{1}{\sqrt{3}}}\left(\dfrac{1}{1-x^4}-1\right)\,\mathrm{d}x\\&=4\,\pi\,\int_{0}^{\frac{1}{\sqrt{3}}}\left(\dfrac{(1-x^2)+(1+x^2)}{2\,\left(1-x^2\right)\,\left(1+x^2\right)}-1\right)\,\mathrm{d}x\\&=4\,\pi\,\int_{0}^{\frac{1}{\sqrt{3}}}\left(\dfrac{1}{2\,\left(1+x^2\right)}+\dfrac{1}{2\,\left(1-x^2\right)}-1\right)\,\mathrm{d}x\\&=4\,\pi\,\int_{0}^{\frac{1}{\sqrt{3}}}\left(\dfrac{1}{2\,\left(1+x^2\right)}+\dfrac{(1-x)+(1+x)}{4\,\left(1-x\right)\,\left(1+x\right)}-1\right)\,\mathrm{d}x\\&=4\,\pi\,\int_{0}^{\frac{1}{\sqrt{3}}}\left(\dfrac{1}{2\,\left(1+x^2\right)}+\dfrac{1}{4\,\left(1+x\right)}+\dfrac{1}{4\,\left(1-x\right)}-1\right)\,\mathrm{d}x\\&=4\,\pi\,\left[\dfrac{\arctan\,x}{2}+\dfrac{1}{4}\,\ln\,\dfrac{1+x}{1-x}-x\right]_{0}^{\frac{1}{\sqrt{3}}}\\&=4\,\pi\,\left(\dfrac{1}{2}\,\arctan\,\dfrac{1}{\sqrt{3}}+\dfrac{1}{4}\,\ln\,\dfrac{\displaystyle{1+\dfrac{1}{\sqrt{3}}}}{\displaystyle{1-\dfrac{1}{\sqrt{3}}}}-\dfrac{1}{\sqrt{3}}\right)\\&=4\,\pi\,\left(\dfrac{\pi}{12}+\dfrac{1}{4}\,\ln\,\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}}\right)\\&=\dfrac{\pi^2}{3}+\pi\,\ln\,\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{4\,\pi}{\sqrt{3}}\end{aligned}}\)

so,

\(\displaystyle{I=\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\dfrac{x^4}{1-x^4}\,\arccos\,\left(\dfrac{2\,x}{1+x^2}\right)\,\mathrm{d}x=\dfrac{\pi^2}{6}+\dfrac{\pi}{2}\,\ln\,\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{2\,\pi}{\sqrt{3}}}\) .
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