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Blazing goIITian

Joined: 11 May 2008

Posts: 491

21 Dec 2009 01:32:55 IST

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$I=\int\frac{e^x dx}{(x+1)^2}-2\int\frac{e^xdx}{(x+1)^3}\\ \\ \\integrating\ by-parts\ to\ first\ part\\ \\ I= \frac{e^x}{(x+1)^2}+2\int\frac{e^xdx}{(x+1)^3}-2\int\frac{e^xdx}{(x+1)^3}+C\\ \\ \\I = \frac{e^x}{(x+1)^2} +C$

Blazing goIITian

Joined: 11 May 2008

Posts: 491

21 Dec 2009 02:06:35 IST

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$put\ x=sin2\theta\\ \\ I=\int ln|(|sin\theta-cos\theta)|+|sin\theta+cos\theta|)|cos2\theta\ d\theta\\ \\ as\ the\ interval\ is \ not\ specified,so\ two\ cases\ arises\\ \\ I=\int ln(2sin\theta)cos2\theta\ d\theta\ (when\ both\ mod\ opens\ with\ same\ sign)\\ \\ I=\int ln(2cos\theta)cos2\theta\ d\theta\ (when\ both \ mod\ opens\ with\ different\ sign)\\ \\ solving\ first\ case,integrating\ by-parts\\ \\ I=ln(2sin\theta).\frac{sin2\theta}{2}-\frac{(\theta+\frac{sin2\theta}{2})}{2}+C\\ \\ put\ back\ \theta\ in\ terms\ of\ x,get\ the\ ans\\ \\ solving\ second \ case,integrating\ by-parts\\ \\ I=ln(2sin\theta).\frac{sin2\theta}{2}+\frac{(\theta-\frac{sin2\theta}{2})}{2}+C\\ \\ put\ back\ \theta\ in\ terms\ of\ x,get\ the\ ans\\ \\$

Blazing goIITian

Joined: 11 May 2008

Posts: 491

21 Dec 2009 02:19:47 IST

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$put\ x=acos\theta\\ \\ I=a^2\ \int_{0}^{\frac{\pi}{2}}\theta^2sin^2\theta\ d\theta\\ \\ or\ I = a^2\ \int_{0}^{\frac{\pi}{2}}cos^2\theta\ (\theta^2+\frac{\pi^2}{4}-\theta\pi)d\theta\\ \\adding\ both\\ \\ I=\frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(\theta^2+\frac{\pi^2}{4}cos^2\theta-\pi\theta\ cos^2\theta)d\theta$

now,all 3 integrals are easily seperately integrable . . .

Blazing goIITian

Joined: 11 May 2008

Posts: 491

21 Dec 2009 02:46:31 IST

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$I=\int_{0}^{1}tan^{-1}\frac{1}{(1-x+x^2)}dx\\ \\ I=\int_{0}^{1}tan^{-1}(\frac{(1-x )+ x}{1-x(1-x)})dx\\ \\ I=\int_{0}^{1}tan^{-1}(1-x)\ dx+\int_{0}^{1}tan^{-1}x\ dx\\ \\ applying\ integration\ b\ parts\ to\ each\\ \\ I=\int_{0}^{1}\frac{1-x}{1+x^2}dx+\frac{\pi}{4}-\int_{0}^{1}\frac{x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-\int_{0}^{1}\frac{2x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-ln2$

kabi

Blazing goIITian

Joined: 11 Jan 2009

Posts: 575

22 Dec 2009 18:24:51 IST

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2. using integration by parts

I = x log (Sqrt(1-x)+sqrt(1+x)) - integration ( ( 1-Sqrt(1-x²)) /( -2Sqrt(1-x²)) dx

= x log (Sq.(1-x)+Sq (1+x)) - 1/2 ( integration ( 1 - (1/Sqrt(1-x² )) dx )

= x log (Sq.(1-x)+Sq (1+x))- x/2 +sin^-1(x)/2 +constant

thank u

Loser!

Blazing goIITian

Joined: 23 Sep 2007

Posts: 966

25 Dec 2009 21:23:35 IST

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Thanx for the replies! I have few more doubts..plz try these

5) $\int cos 2\theta log\left ( \frac{cos\theta+sin\theta }{cos\theta -sin\theta } \right )d\theta$

6) $ST \int e^{tan^{-1}x}\left ( \frac{1+x+x^2}{1+x^{2}} \right )dx=xe^{tan^{-1}x}+c$

7) $\int xe^{x}cosx dx$

8) $\int log(1+cosx)-xtan(\frac{x}{2})$

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

25 Dec 2009 21:47:02 IST

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first can be done by parts

= cos2theta * log [tan(pie/4 +theta)]

=1/2*log [tan(pie/4 +theta)]* sin2theta - integral of 1/ tan(pie/4 +theta) * sec^2 (pie/4 +theta ) *sin2theta /2

after simplification of second integrand we get

=1/2*log [tan(pie/4 +theta)]* sin2theta - inetgral of 2tan2theta

{ bcoz 1/ tan(pie/4 +theta) * sec^2 (pie/4 +theta ) = 1/ sin(pie/4 +theta)cos(pie/4 +theta) = 2/sin2(pie/4 +theta) = 2/ sin(pie/2 +2theta) = 2/ cos2theta which is in multiplication with sin2theta so becomes tan2theta}

=1/2*log [tan(pie/4 +theta)]* sin2theta - log mod sec2theta

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

25 Dec 2009 21:49:37 IST

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= e^tan inv x { 1 + x/(1+x^2 )}dx

substitute tan invx x =y

= e^y { 1+ tany /sec^2 y } sec^2 y dy

= e^y { sec^2 y +tany }dy

=e^y tany (integration by parts )

now resubsitute

= x e^tan inv x

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

25 Dec 2009 21:51:17 IST

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7> this one's is easy

just take x as first function and e^x cosx as second function

then integrate

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

25 Dec 2009 21:54:11 IST

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edited

Akshay

Hot goIITian

Joined: 25 Jun 2008

Posts: 174

25 Dec 2009 22:32:49 IST

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8) The reply given above is wrong as u cant solve further

Instead use by parts directly keeping the second integral as it is.

x.log[1 + cosx] - ∫x.(-sinx)/(1 + cosx)dx - ∫xtan(x/2)dx

x.log[1 + cosx] + ∫x. [ 2 sin(x/2)cos(x/2)]/[2cos^2 (x/2)]dx - ∫xtan(x/2)dx

xlog[1 + cosx] +∫xtan(x/2)dx - ∫xtan(x/2)dx

xlog[1 + cos(x/2)] + c

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

25 Dec 2009 22:38:22 IST

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oops sorry ...thnks for pointing it out

Blazing goIITian

Joined: 28 Oct 2009

Posts: 594

25 Dec 2009 23:03:05 IST

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edited

Blazing goIITian

Joined: 28 Oct 2009

Posts: 594

25 Dec 2009 23:03:37 IST

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edited

Blazing goIITian

Joined: 28 Oct 2009

Posts: 594

25 Dec 2009 23:19:13 IST

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jus take x=cos2y and then use by parts twice

Akshay

Hot goIITian

Joined: 25 Jun 2008

Posts: 174

26 Dec 2009 00:10:27 IST

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7) Integrating by parts

Taking x as first function and e^x cosx as second function

x ∫cosx e^xdx - ∫ ∫cosx e^xdxdx -----(i)

Now solving ∫e^x(cosx)dx

Let I = ∫e^x(cosx)dx

solving by parts let cosx be first function

I = e^x(cosx) - ∫e^x(-sinx)dx

I = e^x(cosx) + ∫e^x(sinx)dx

Again byparts sinx is first function

I = e^x(cosx) + e^x(sinx) - ∫e^x(cosx)dx

I = e^x(cosx +sinx) - I

2I = e^x(cosx +sinx)

I = e^x(cosx +sinx)/2

Now substituting in integral (i)

xe^x(cosx + sinx)/2 -∫e^x(cosx +sinx)/2dx

xe^x(cosx +sinx)/2 - e^x(sinx)/2 + c [ By using the formula ∫e^x{f(x) + f'(x)}dx = e^x(f(x))]