integrate pl help me

[satishinamdar]

Dear Sir,
Can you please help me solve the following?

Integral 0 to a dx/{x +sqrt(a sq-x sq)}

Is there any rule or tips to solve integration problems?
Regards

 

[Gene]

I would start by multiplying top and bottom by x-sqrt(a²-x²).
You should recognize that sqrt(a²-x²) is the equation of the top half of a circle of radius a and the integral 0 to a is a quarter of a circle.

 

[satishinamdar]

Dear Sir,
Thanks for reply.However unable to understand.Please make it simple.The book gives answer as Pi/4.

Also pl let me know some tips to solve integrations.

Regards

 

[pka]

\(\displaystyle \L
\frac{1}{{x + \sqrt {a^2 - x^2 } }}\left( {\frac{{x - \sqrt {a^2 - x^2 } }}{{x - \sqrt {a^2 - x^2 } }}} \right) = \frac{-1}{{a^2 }}\left( {x - \sqrt {a^2 - x^2 } } \right)\)

\(\displaystyle \L
\int {\left( { \sqrt {a^2 - x^2 } } \right)dx = \frac{{\left( {x\sqrt {a^2 - x^2 } } \right)}}{2} + \left( {\frac{{a^2 }}{2}} \right)\arcsin \left( {\frac{x}{a}} \right)}\)

 

[soroban]

Hello, satishinamdar!

Am I missing something?
. . There is no easy way to integrate this . . .

\(\displaystyle \L\frac{1}{{x\,+\,\sqrt {a^2 - x^2 } }}\cdot\left({\frac{{x\,-\,\sqrt{a^2\,-\,x^2}}}{{x\,-\,\sqrt {a^2-x^2 }}}}\right)\:=\:\frac{-1}{{a^2 }}\left( {x\,-\,\sqrt{a^2 - x^2}} \right)\) . This is incorrect!

The denominator is: \(\displaystyle \x\,+\,\sqrt{a^2\,-\,x^2})(x\,-\,\sqrt{a^2\,-\,x^2})\:=\:x^2\,-\,(a^2\,-\,x^2)\:=\:2x^2\,-\,a^2\)


The integral is far more difficult than you think . . .

Let $\displaystyle x\,=\,a\cdot\sin\theta\;\;\Rightarrow\;\;dx\,=\,a\cdot\cos\theta\,d\theta$

Note that: $\displaystyle \sqrt{a^2\,-\,x^2}\:=\:\sqrt{a^2\,-\,a^2\sin^2\theta}\:=\:\sqrt{a^2(1\,-\,\sin^2\theta)}\:=\:\sqrt{a^2\cos^2\theta}\:=\:a\cdot\cos\theta$

Substitute: \(\displaystyle \L\:\int\frac{a\cdot\cos\theta\,d\theta}{a\cdot\sin\theta\,+\,a\cdot\cos\theta}\;= \;\int\frac{\cos\theta}{\sin\theta\,+\,\cos\theta}\,d\theta\)

Divide top and bottom by \(\displaystyle \cos\theta:\;\;\L\int\frac{d\theta}{\tan\theta\,+\,1}\)

Let $\displaystyle v\,=\,\tan\theta\;\;\Rightarrow\;\;\theta\,=\,\arctan v\;\;\Rightarrow\;\;d\theta\,=\,\frac{dv}{v^2\,+\,1}$

Substitute: \(\displaystyle \L\:\int\frac{\left(\frac{dv}{v^2\,+\,1}\right)}{v\,+\,1}\;=\;\int\frac{dv}{(v\,+\,1)(v^2\,+\,1)}\)

Partial fractions: \(\displaystyle \L\:\frac{1}{(v\,+\,1)(v^2\,+\,1)}\;=\;\frac{A}{v\,+\,1}\,+\,\frac{Bv\,+\,C}{v^2\,+\,1}\)

. . and we get: $\displaystyle \,A\,=\,\frac{1}{2},\:B\,=\,-\frac{1}{2},\:C\,=\,\frac{1}{2}$

The integral becomes: \(\displaystyle \L\:\int\left(\frac{\frac{1}{2}}{v\,+\,1}\,+\,\frac{-\frac{1}{2}v}{v^2\,+\,1}\,+\,\frac{\frac{1}{2}}{v^2\,+\,1}\right)\,dv\)

. . . \(\displaystyle \L=\;\frac{1}{2}\left[\int\frac{dv}{v\,+\,1}\,-\,\int\frac{v\,dv}{v^2\,+\,1}\,+\,\int\frac{dv}{v^2\,+\,1}\right)\)

. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln|v\,+\,1|\,-\,\frac{1}{2}\ln(v^2\,+\,1)\,+\,\arctan v\right]\)

Multiply by \(\displaystyle \frac{2}{2}:\L\;\;\frac{1}{4}\left[2\cdot\ln|v\,+\,1|\,-\,\ln(v^2\,+\,1)\,+\,2\cdot\arctan v\right]\)

. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln(v\,+\,1)^2 - \ln(v^2\,+\,1) + 2\cdot\arctan v\right]\)

. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(v\,+\,1)^2}{v^2\,+\,1}\,+\,2\cdot\arctan v\right]\)


Back-substitute: \(\displaystyle \L\:\frac{1}{4}\left[\ln\frac{(\tan\theta\,+\,1)^2}{\tan^2\theta\,+\,1}\,+\,2\cdot\arctan(\tan\theta)\right]\)

. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(\tan\theta\,+\,1)^2}{\sec^2\theta}\,+\,2\theta\right]\)


Back-substitute: Since $\displaystyle \sin\theta\,=\,\frac{x}{a}\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{x}{\sqrt{a^2-x^2}},\:\sec\theta\,=\,\frac{a}{\sqrt{a^2-x^2}}$

And we have: \(\displaystyle \L\:\frac{1}{4}\left[\ln\frac{\left(\frac{x}{\sqrt{a^2-x^2}}\,+\,1\right)^2}{\left(\frac{a}{\sqrt{a^2-x^2}}\right)^2}\,+\,2\cdot\arcsin\left(\frac{x}{a}\right)\right]\)

. . . \(\displaystyle \L=\;\frac{1}{4}\left[\ln\frac{(x\,+\,\sqrt{a^2-x^2})^2}{a^2}\,+\,2\cdot\arcsin\left(\frac{x}{a}\right)\right]\)

. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln\frac{x\,+\,\sqrt{a^2-x^2}}{a}\,+\,\arcsin\left(\frac{x}{a}\right)\right]^a_0\)


We have: \(\displaystyle \L\:\frac{1}{2}\left[\ln\left(\frac{a\,+\,\sqrt{0}}{a}\right)\,+\,\arcsin\left(\frac{a}{a}\right)\right]\,-\,\frac{1}{2}\left[\ln\left(\frac{0\,+\,\sqrt{a^2}}{a}\right)\,+\,\arcsin\left(\frac{0}{a}\right)\right]\)

. . . \(\displaystyle \L=\;\frac{1}{2}\left[\ln(1)\,+\,\arcsin(1)\right]\,-\,\frac{1}{2}\left[\ln(1)\,+\,\arcsin0\right]\)

. . . \(\displaystyle \L=\;\frac{1}{2}\left[0\,+\,\frac{\pi}{2}\right]\,-\,\frac{1}{2}\left[0\,+\,0\right}\)

. . . \(\displaystyle \L=\;\frac{\pi}{4}\) . . . . There!

 

[galactus]

Soroban, you beat me to it. After all that work, I am posting anyway :lol: .

This is a booger of an integral no matter how you look at it.

\(\displaystyle \int{\frac{dx}{x+sqrt{a^{2}-x^{2}}}\)

Let $\displaystyle x=(a)sin{\theta}; dx=(a)cos{\theta}d{\theta}$

$\displaystyle =\frac{(a)cos{\theta}}{(a)sin{\theta}+sqrt{a^{2}-((a)sin{\theta})^{2}}}$

$\displaystyle =\frac{(a)cos{\theta}}{(a)sin{\theta}+(a)cos{\theta}}$

$\displaystyle =\int{\frac{cos(x)}{cos(x)+sin(x)}}dx$

$\displaystyle =\int{\frac{1}{tan(x)+1}}dx$

Let $\displaystyle u=tan(x); x=tan^{-1}(u), dx=\frac{1}{u^{2}+1}$

$\displaystyle \int{\frac{1}{u^{3}+u+u^{2}+1}}du$

$\displaystyle =\frac{1}{2}\int{\frac{-u+1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

$\displaystyle =\frac{1}{2}\int{\frac{-u+1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

$\displaystyle =\frac{-1}{2}\int{\frac{u}{u^{2}+1}}du+\frac{1}{2}\int{\frac{1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

Let $\displaystyle u_{1}=u^{2}+1; du_{1}=2udu; \frac{du_{1}}{2}=udu$

$\displaystyle \frac{-1}{2}\int{\frac{1}{2u_{1}}}du_{1}+\frac{1}{2}\int{\frac{1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

\(\displaystyle =\frac{-1}{4}\int{\frac{1}{u_{1}}du_{1}+\frac{1}{2}\int{\frac{1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du\)

$\displaystyle =\frac{-1}{4}ln(u_{1})+\frac{1}{2}\int{\frac{1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

$\displaystyle =\frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}\int{\frac{1}{u^{2}+1}}du+\int{\frac{1}{2u+2}}du$

Let $\displaystyle u=tan(u_{1}), du=sec^{2}(u_{1})du_{1}$

$\displaystyle \frac{sec^{2}(u_{1})du_{1}}{tan^{2}(u_{1})+1}=1du_{1}$

$\displaystyle =\frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}\int{du_{1}}+\int{\frac{1}{2u+2}}du$

$\displaystyle =\frac{-1}{4}(u^{2}+1)+\frac{1}{2}u_{1}+\int{\frac{1}{2u+2}}du$

$\displaystyle =\frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}tan^{-1}(u)+\int{\frac{1}{2u+2}}du$

Let $\displaystyle u_{1}=u+1; du_{1}=du$

\(\displaystyle \frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}tan^{-1}(u)+\frac{1}{2}\int{\frac{1}{u_{1}}du\)

$\displaystyle =\frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}tan^{-1}(u)+\frac{1}{2}ln(u_{1})$

$\displaystyle =\frac{-1}{4}ln(u^{2}+1)+\frac{1}{2}tan^{-1}(u)+\frac{1}{2}ln(u+1)$

$\displaystyle =\frac{-1}{4}ln(1+tan^{2}(x))+\frac{1}{2}tan^{-1}(tan(x))+\frac{1}{2}ln(tan(x)+1)$

Let $\displaystyle tan{\theta}=\frac{x}{sqrt{a^{2}-x^{2}}}$

Everything becomes 0 except for \(\displaystyle \frac{1}{2}tan^{-1}\frac{x}{sqrt{a^{2}-x^{2}}\)

=$\displaystyle \frac{sin^{-1}(\frac{x}{a})}{2}$

Let x=a and we have:

$\displaystyle \frac{sin^{-1}(1)}{2}=\frac{{\pi}}{4}$

WHEW!!!........DAMN!!!........I'M GONNA GO HAVE A DRINK!

 

[satishinamdar]

Thank you for replies.
This is an example for class 11.(Where calculas has just started.
My son who is in 11 std had this proble.
Subsequently he got some solution like below.
he put x=asin theta
changed the limits from 0 to a to
0 to pi/2
then he proceeded and got answer in 4 to 5 steps.
I am not convinced .Has he assumed something and proceeded?
Can you try any other simple methos and reply.
I will be greatful.
How to use signs of integration?Write theta etc?
Rggards.
I hope you have got what my son did.If you want to knw how he proceeded it is as follows

put x=asintheta
dx=acostheta
change the limits
int 0 to pi/2 acos theta/asintheta+sqrt (asq-asq sinsqtheta
int 0 to pi/2 acos theta (pi/2-theta)dtheta/ asin(pi/2-theta)+sqrt(asq-asqsinsqtheta(pi/2-theta)
int 0 to pi/2 asintheta/acostheta+sqrt(asqcossqtheta)-----


pl help
anyway he has solved but is the process correct?

 

[soroban]

Hello, satishinamdar!

I tried to follow what your son did . . . but failed . . .

We have: \(\displaystyle \L\:\int^{\;\;\;a}_0\frac{dx}{x\,+\,\sqrt{a^2\,-\,x^2}}\)

and he did the following:

Put $\displaystyle x\,=\,a\cdot\sin\theta\;\;\Rightarrow\;\;dx\,=\,a\cdot\cos\theta\cdot d\theta$

Change the limits: \(\displaystyle \L\;\int^{\;\;\;\frac{\pi}{2}}_0 \frac{a\cdot\cos\theta\cdot d\theta}{a\cdot\sin\theta\,+\,\sqrt{a^2\,-\,a^2\sin^2\theta}}\)

\(\displaystyle \L\;\;\;=\;\int^{\;\;\;\frac{\pi}{2}}_0 \frac{a\cdot\cos\theta(\overbrace{\frac{\pi}{2}\,-\,\theta})d\theta}{a\cdot\sin(\underbrace{\frac{\pi}{2}\,-\.\theta})\,+\,\sqrt{a^2\,-\,a^2\sin^2(\underbrace{\frac{\pi}{2}\,-\,\theta})} }\)
. . . . . . . . . . . . . . . . . ??

I don't understand why he tried to change to the <u>complement</u> of the angle . . . incorrectly.

If he replaces $\displaystyle \theta$ with $\displaystyle \frac{\pi}{2}\,-\,\theta$, he must also change the <u>limits</u>.

Besides, he must also change the <u>function</u> since: $\displaystyle \,\sin\theta\,=\,\cos\left(\frac{\pi}{2} - \theta\right)$ and $\displaystyle \ \cos\theta\,=\,\sin\left(\frac{\pi}{2} - \theta\right)$

. . That is: $\displaystyle \,\sin\theta\,\neq\,\sin\left(\frac{\pi}{2}\,-\,\theta\right)$

 

[satishinamdar]

Thanks
There was a mistake from my side.sorry.
I should not have written acos theta (pi/2-theta)
It should have been a cos(pi/2-theta).

can it be solved then?
Regards

 

[galactus]

This integral isn't pretty no matter how you look at it. If someone knows a 'simple' method, I would be interested in seeing it myself.

 

[satishinamdar]

Dear Sir,
Anyway Thanks a lot.