Home » Ask & Discuss » Mathematics. » Integral Calculus « Back to Discussion

Integral Calculus

Blazing goIITian

Joined:	23 Sep 2007
Post:	966

1 Jan 2010 21:36:02 IST

0 People liked this

2402

Loading..Definte integration

None

1) $\int_{-3}^{-1}(x+1)(x+2)(x+3)dx$

2)If $\int_{0}^{\pi /2}cos^{m}xsin^{m}xdx=\lambda \int_{0}^{\pi /2}sin^{m}xdx$ then Prove that $\lambda =2^{-m}$

3) $\int_{0}^{2a}x^{3}\sqrt{2ax-x^{3}}=\frac{7}{8}\pi a^{5}$

4)Let $f(x)=\int_{1}^{x}\sqrt{2-t^{2}}dt$ . Then find the real roots of the equation $x^{2}-f^{'}(x)=0$ ?

Share this article on:

Comments (33)

Go to Page...

kabi

Blazing goIITian

Joined: 11 Jan 2009

Posts: 575

1 Jan 2010 22:01:19 IST

Like 1 people liked this

i m in a hurry so just giving answer to 1st one .

int ( -3 to -1 ) (x+1)(x+2)(x+3) dx

Transformation x+2 =y

=int (-1 to 1 ) (y-1) y (y+1)dy

= 0

eragon24

Blazing goIITian

Joined: 14 May 2009

Posts: 697

1 Jan 2010 22:53:59 IST

Like 2 people liked this

eragon24

Blazing goIITian

Joined: 14 May 2009

Posts: 697

1 Jan 2010 22:56:26 IST

Like 0 people liked this

edited

Deepak Aggarwal

Blazing goIITian

Joined: 9 May 2009

Posts: 766

2 Jan 2010 18:19:34 IST

Like 4 people liked this

4. Apply lebnitz theorem,

So, u will get f'(x) = ( 2 - x² )^1/2

Since, x² = f'(x) = ( 2 - x² )^1/2

So, x⁴ = 2 - x²

or, x⁴ + x² - 2 = 0

or, x⁴ + 2x² - x² - 2 = 0

or, (x² - 1) (x² + 2) = 0

Now since x is real, so x² = 1 and other value is neglected

therefore, x = + 1 which are the required roots

Deepak Aggarwal

Blazing goIITian

Joined: 9 May 2009

Posts: 766

2 Jan 2010 18:23:14 IST

Like 0 people liked this

3. Try solving integral with the properties of definite integral.

U will get (4096)a^11/2/ (3465) = (7/8)(pi)a⁵

Solving it, u get a = ( 33 (35)^1/2 ) / 128

Calculation is a bit complicated in this question.

I would try it later if it could be solved by some other method.

taran

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

3 Jan 2010 01:06:37 IST

Like 0 people liked this

in 3) try putting x =square root of (2a)*sintheta

Loser!

Blazing goIITian

Joined: 23 Sep 2007

Posts: 966

4 Jan 2010 15:37:33 IST

Like 0 people liked this

thank a ton for helping me out.. can u plz post post d complete solution for 3rd one..thnx.. also few more doubts

5)If $y(x)=\int_{\frac{\pi ^{2}}{16}}^{x^{2}}\frac{cosx.cos\sqrt{\theta }}{1+sin^{2}\sqrt{\theta }}d\theta$ then find $\frac{dy}{dx}$ at $x=\pi$

6)Prove that $\int_{0}^{1}\frac{logx}{\sqrt{1-x^{2}}}dx=\frac{\pi }{2}.log\frac{1}{2}$

7)Show that $\int_{0}^{\frac{\pi }{2}}sinxlog(sinx)dx=\log_{e}(\frac{2}{e})$

8) $\int_{0}^{\pi }\frac{xdx}{1+cos^{2}x}=\frac{\pi ^{2}}{2\sqrt{2}}$

yamini

Hot goIITian

Joined: 2 Apr 2009

Posts: 161

4 Jan 2010 18:14:43 IST

Like 0 people liked this

I = integral 0-pi xdx/1+cos^2x .........1eqn

I=integral 0-pi (pi-x)dx/1+cos^2(pi-x) = (dx/1+cos^2xpi-x)....2eqn

adding 1 and 2

2I = integral 0-pi pidx/1+cos^2x

I = pi/2 integral 0-pi dx/1+cos^2x = pi/2(0-pi dx/1+cos^2x +0-pi/2 dx/1+cos^2(pi-x))

=pi integral 0-pi sec^2xdx/ 2+tan^2 x .........put tanx = t

= pi integral 0-infinity dt/t^2+2 =pi/root2 tan inverse (t/root2) limits 0- infinity

=pi^2/2root2

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

4 Jan 2010 18:23:43 IST

Like 1 people liked this

Differentiate using Newton Leibnitz theorem,

f'(x) = 2xcos²x/(1+sin²(x))

f'(pi) = 2(pi)

6) (Not able to post using goiit equation editor so i'll just give you the idea)

Let sin(t) = x

Then t = sin^-1(x)

i.e. dt = dx/{1-x²}^1/2

Therefore the integral reduces to INTEGRAL (0 to pi/2) {log(sint)dt}

Let I = INTEGRAL (0 to pi/2) log(sint)dt

i.e. I = INTEGRAL (0 to pi/2) log(cost)dt

so 2I = INTEGRAL (0 to pi/2) log(sint.cost)dt

= INTEGRAL (0 to pi/2) log(sin2t/2)dt

= INTEGRAL (0 to pi/2)logsin2tdt - INTEGRAL(0 to pi/2)log2dt

But , INTEGRAL (0 to pi/2) log(sin2t)dt = INTEGRAL (0 to pi)(0.5sinzdz) = I

i.e. I = -(pi/2)log(2)

8) Is easy. Use the property INTEGRAL(0 to a) f(x)dx = INTEGRAL (0 to a)f(a-x)dx

So the integral reduces to I = INTEGRAL (0 to pi)0.5(pi)dx/{1+cos²x}

Now, replace cosx = 1/secx. However this makes the function discontinuous at x=(pi/2)

So split the integral as INTEGRAL (0 to pi/2) + INTEGRAL (pi/2 to pi)

Now replace cosx = 1/secx, put tanx = t and proceed.

yamini

Hot goIITian

Joined: 2 Apr 2009

Posts: 161

4 Jan 2010 22:32:53 IST

Like 3 people liked this

yamini

Hot goIITian

Joined: 2 Apr 2009

Posts: 161

4 Jan 2010 22:33:28 IST

Like 3 people liked this

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

4 Jan 2010 23:03:29 IST

Like 1 people liked this

$I = \int_{0}^{\pi/2}\log \sin(x)dx$ I =

$I = \int_{0}^{\pi/2}\log \cos(x)dx$ (integral property)

2I = $\int_{0}^{\pi/2}\log( \sin(x)cos(x))dx$

since 2sinxcosx =sin2x,

$\int_{0}^{\pi/2}\log( \sin(2x)/2)dx$ = 2I

$2I = \int_{0}^{\pi/2}\log( \sin(2x))dx-\int_{0}^{\pi/2}\log(2)dx$

Consider,

$I1 = \int_{0}^{\pi/2}\log(\sin(2x))dx$

Put 2x = t

So you have

$I1 = 0.5\int_{0}^{\pi}\log(\sin(t))dt$

Use property of integrals,

$I1 = 0.5\int_{0}^{\pi/2}\log(\sin(t))dt +0.5\int_{\pi/2}^{\pi}\ln(\sin(t))dt$

Now you can easily prove that I1 = I

hence I = -(pi/2) (log2)

=(pi/2)(log(1/2))

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

4 Jan 2010 23:36:37 IST

Like 1 people liked this

This obviously follows from ,

$I = \int_{0}^{1}\log(x)dx/\sqrt{(1-x^2)}$

Let $\sin^{-1}(x) = t$

so $I = \int_{0}^{\pi/2}\log\sin(t)dt = \int_{0}^{\pi/2}\log\sin(x)dx$

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

5 Jan 2010 00:20:09 IST

Like 3 people liked this

$\int_{0}^{\pi/2}\log\sin(x)dx + \int_{\pi/2}^{\pi}\log\sin(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{\pi/2-\pi/2}^{\pi-\pi/2}\log\sin(x+\pi/2)dx$

$= \int_{0}^{\pi/2}\log\sin(x)dx + \int_{0}^{\pi/2}\log\cos(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{0}^{\pi/2}\log\sin(x)dx$

$= 2\int_{0}^{\pi/2}\log\sin(x)dx$

BUT

$I1 = (0.5)2\int_{0}^{\pi/2}\log\sin(x)dx$

Hence,

$I1 = \int_{0}^{\pi/2}\log\sin(x)dx = I$

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

5 Jan 2010 21:24:28 IST

Like 2 people liked this

$\int_{0}^{\pi/2}(log\sin(x))\sin(x)dx = I$

INTEGRATE BY PARTS, EVAULATE THE INDEFINITE INTEGRAL FIRST

$\int \log\sin(x)\sin(x)dx = \log\sin(x)\int \sin(x)dx - \int [\frac{\mathrm{d} \log\sin(x)}{\mathrm{d} x}\int \sin(x)dx]dx$

$= -\log\sin(x)\cos(x) + \int [\cos^2(x)/\sin(x)]dx$

$= -\log\sin(x)\cos(x) + \int [\csc(x) - \sin(x)]dx$

$= -\log\sin(x)\cos(x) + \int \csc(x)dx - \int\sin(x)dx$

$= -\log\sin(x)\cos(x) + \log\tan(x/2) + \cos(x)dx$ (TYPO HERE, IGNORE 'dx')

NOW APPLY THE LIMITS,

$-\log\sin(\pi/2)\cos(\pi/2) + \log\tan(\pi/2/2) + \cos(\pi/2) = 0$ .........(1)

NOW FOR THE LOWER LIMIT, AS THE LOG FUNCTIONS BECOME INFINITE AT X=0, WE HAVE TO CONSIDER THE LIMITING VALUE

SO, LOWER LIMIT

$= \lim_{x\rightarrow0}\log(\tan(x/2)/\sin^{\cos(x)}(x)) + \cos(0)$

$= \log(\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x)))+ \cos(0)$

CONSIDER

$\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x))$

EVALUATE THIS LIMIT BY L'HOSPITAL RULE,

$\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x)) =-\log(2)$

SO THE LOWER LIMIT BECOMES

$=-\log(2)+ 1 = -\log(2) + \log(e) = \log(e/2)$ ................(2)

NOW ,

$I = (1)-(2) = -\log(e/2) = \log(2/e)$

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

5 Jan 2010 21:28:18 IST

Like 0 people liked this

$\log$ denotes logarithm to the base "e".

Yagyadutt Mishra

Forum Expert

Joined: 19 Feb 2009

Posts: 1984

6 Jan 2010 16:25:37 IST

Like 1 people liked this

Yagyadutt Mishra

Forum Expert

Joined: 19 Feb 2009

Posts: 1984

6 Jan 2010 16:34:31 IST

Like 1 people liked this

The third question will give rise to a hypergeometric function........

so dont worry for that one...

Akshay

Blazing goIITian

Joined: 27 Dec 2007

Posts: 689

6 Jan 2010 18:52:45 IST

Like 2 people liked this

For the question 7, I got a much simpler method.

Put cosx = t , you have,

Integrate by parts,

Take 1 as the function u, and the log term as function v,

$\int \log\sqrt{1-t^2}dt = t\log\sqrt{1-t^2} - \int [\frac{\mathrm{d}\log\sqrt{1-t^2} }{\mathrm{d} x}\int dt]dt$

(differentiation is with respect to "t" and not x )

$\int \log\sqrt{1-t^2}dt = t\log\sqrt{1-t^2} + \int t^2dt/[1-t^2]$

$\int t^2dt/[1-t^2] = \int -dt + \int dt/(1-t^2)$

$\int t^2dt/[1-t^2] = -t + \log(\sqrt{1-t/1+t})$ (theres a typo here, The signs are interchanged within the log term it's actually sqrt(1+t/1-t))

So $\int\log\sqrt{[1-t^2]}dt= -t + \log(\sqrt{1-t/1+t}) + t\log(\sqrt{1+t})+t\log(\sqrt{1-t})$

$\int\log\sqrt{[1-t^2]}dt= -t + (t+1)\log(\sqrt{1+t})+(t-1)\log(\sqrt{1-t})$

Limits from 0 to 1,

$\lim_{x\rightarrow1}(t-1)\log(\sqrt{1-t}) = \lim_{x\rightarrow1}(\log(\sqrt{1-t}/(1/[1-t]))$ (This is a $0.\infty$ form converted into 0/0. Evaluate using L'Hospital Rule)

You will see that this value comes out to be zero

Hence the upper limit value comes out to be $-1 + (1+1)\log(\sqrt{1+1})+ \lim_{x\to 1}(t-1)\log(\sqrt{1-t}) = -1 + 2\log(\sqrt{2}) + 0$

Lower limit t=0 = 0

Hence the value of the integral

$\int_{0}^{\pi/2}\sin(x)\log\sin(x)dx = \int_{0}^{1}\log(\sqrt{1-t^2})dt = \log(2/e)$

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team