integral THREAD.

thnks yaar.
Bohat time laga post karne main.........
soch raha tha pura post nahin karun magar karte karte pura hi kar diya [3][3]

ans to q12....(2n+1)pi.....

in the q we did not assume n ti be odd or even....it is for all n..............
hence the answer will be applicable here also.......

i mean we are only putting specific values of n......suppose we need to find it for n=3...we kno for general n..just plug in the value of n to get the answer.........

yup i know that but can that method used der can be applied to this question 12 becoz it's slighty different [7]

oops yes corrected...

\int_{0}^{\pi /2}{log(\frac{1+tan^{2}\theta}{tan\theta }) }d\theta = \int_{0}^{\pi /2}-2logcos\theta + \int_{0}^{\pi /2}logtan\theta

and \int_{0}^{\pi /2}logtan\theta d\theta = 0

@ carl

putting x=tanθ
we get

\int log\left( \frac{2}{sin2\theta}\right)d\theta

correct me if i am wrong

\int \left(cos( blog\left(\frac {x}{a} \right)\right)

put \frac{x}{a} =u \rightarrow \frac{1}{a} dx = du

\textsl{integral reduces to}

\frac{1}{a} \int cos\left(b\left(log u \right) \right) du

\textrm{log u = t } \rightarrow e^{t} = u

\frac{1}{u} du = dt

\texttt{integral reduces to }

\frac{1}{a}\int u.cos( bt )

\frac{1}{a}\int e^t.cos( bt )

this is of the form

\int e^ {ax} .cos ( bx ) dx

here a= 1 and x=t

now we have a standard result for this obtained from by parts

put a=1 and u get the answer [1]

I =\int \frac{dx}{( asec x + sinx )^2}

I =\int \frac{cos^2x .dx}{( a + sinxcosx )^2}

I =\int \frac{cos^2x .dx}{( a^2 + 2a sinx cosx + sin^2x .cos^2x )}

I =\int \frac{cos^2x .dx}{( a^2 + a .sin2x+ \frac{1}{4}.sin^2(2x) )}

I =\int \frac{4cos^2x .dx}{( 4a^2 + 4a .sin2x+ .sin^2(2x) )}

I =2\int \frac{(1 + cos2x) .dx}{( 2a + sin2x )^2}

I =2\int \frac{ dx}{( 2a + sin2x )^2}+ 2\int \frac{cos2x}{( 2a + sin2x )^2}

I =2I_{1} + \int \frac{dt}{t^2} \left \left\{ ( \rightarrow 2a +sin2x = t \right\right\}

I =2I_{1} -\frac{1}{( 2a + sin2x )^2}

now evaluatin I₁

A = \frac{cos2x}{2a + sin2x }

\frac{\partial A}{\partial x} = \frac{-4a sin2x-2 }{( 2a + sin2x )^2}

\frac{\partial A}{\partial x} = \frac{-4a( sin2x + 2a) -2 + 8a^2 }{( 2a + sin2x )^2}

\frac{\partial A}{\partial x} = \frac{-4a}{2a + sin2x } + \frac{8a^2 -2 }{(2a+sin2x)^2}

integratin both sides wrt x

A = \int \frac{-4a}{2a + sin2x } +\int \frac{8a^2 -2 }{(2a+sin2x)^2}

\int cosh( logx )

far practize

Convert cosθ into half angles of tan then substitue tanθ/2 = t ..and then it's easily integrable..
\int \sqrt{\frac{(1-cos\theta )}{cos\theta (1+cos\theta )(2+cos\theta )}}d\theta = \int sec^{2}\frac{\theta }{2}\sqrt{\frac{(1 + tan^{2}\frac{\theta}{2} -(1-tan^{2}\frac{\theta}{2})}{1-tan^{2}\frac{\theta}{2}(1+tan^{2}\frac{\theta }{2}+1-tan^{2}\frac{\theta}{2})(2+2tan^{2}\frac{\theta}{2}+1-tan^{2}\frac{\theta}{2})}}d\theta

Now put tanθ/2 = t ..then u will get

= \int \sqrt{\frac{ 1 }{(1-t^{2})(3+2t^{2})}}2tdt

Now put t² = z..
and then the remaining part is quite easy..completing the square and then using the formula...

And for those who cant see the first step clearly..u can use this link..

http://latex.codecogs.com/gif.latex?\int%20\sqrt{\frac{(1-cos\theta%20)}{cos\theta%20(1+cos\theta%20)(2+cos\theta%20)}}d\theta%20=%20\int%20sec^{2}\frac{\theta%20}{2}\sqrt{\frac{(1%20+%20tan^{2}\frac{\theta}{2}%20-(1-tan^{2}\frac{\theta}{2})}{1-tan^{2}\frac{\theta}{2}(1+tan^{2}\frac{\theta%20}{2}+1-tan^{2}\frac{\theta}{2})(2+2tan^{2}\frac{\theta}{2}+1-tan^{2}\frac{\theta}{2})}}d\theta

since no one trying Q7) i'm posting it. I got the answer to that.continued from above post....

\int_{0}^{3\pi /n}{\frac{4(x+3n\pi )}{\left[(a^{2}+b^{2})+(a^{2}-b^{2})cos \frac{2n}{3}(x+3n\pi ) \right]^{2}}dx}
\int_{0}^{3\pi /n}{\frac{4(x+3n\pi )}{\left[(a^{2}+b^{2})+(a^{2}-b^{2})cos \frac{2nx}{3} \right]^{2}}}dx
\frac{3}{n}\int_{0}^{\pi }{\frac{4(\frac{3}{n}x+3n\pi )}{\left[(a^{2}+b^{2})+(a^{2}-b^{2})cos 2x \right]^{2}}}dx ( BY PROPERTY )
Put cos2x = cos²x - sin²x and write (a²+b²) as (a²+b²)(cos²x+sin²x) and after substituting expand and u will get

\frac{12}{n}\int_{0}^{\pi }{\frac{(\frac{3}{n}x+3n\pi )}{\left(2a^{2}cos^{2}x+2b^{2}sin^{2}x\right)^{2}}}dx
I= \frac{9}{n^{2}}\int_{0}^{\pi }{\frac{x+n^{2}\pi }{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}dx..........(1)
I= \frac{9}{n^{2}}\int_{0}^{\pi }{\frac{\pi -x+n^{2}\pi }{\left(a^{2}cos^{2}(\pi -x)+b^{2}sin^{2}(\pi -x)\right)^{2}}}dx
I= \frac{9}{n^{2}}\int_{0}^{\pi }{\frac{\pi -x+n^{2}\pi }{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}dx ...............(2)
adding (1) and (2) we get
2I = \frac{9}{n^{2}}\int_{0}^{\pi }{\frac{(2n^{2}+1)\pi }{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}dx

now 2I = \frac{9(2n^{2}+1)\pi }{n^{2}}\int_{0}^{\pi }{\frac{dx}{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}
= \frac{9(2n^{2}+1)\pi }{n^{2}}.2\int_{0}^{\pi/2 }{\frac{dx}{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}.....( by property )
I = \frac{9(2n^{2}+1)\pi }{n^{2}}\int_{0}^{\pi/2 }{\frac{dx}{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)^{2}}}
Now let I₁ = \int_{0}^{\pi/2 }{\frac{dx}{\left(a^{2}cos^{2}x+b^{2}sin^{2}x\right)}}
= \int_{0}^{\pi/2 }{\frac{sec^{2}xdx}{\left(a^{2}+(btanx)^{2}\right)}}

put b tanx = t → sec^{2}x dx = \frac{dt}{b}

I₁ = \frac{1}{b}\int_{0}^{\infty }{\frac{dt}{a^{2}+t^{2}}} = \frac{1}{ab}\left\{tan^{-1}(\infty )-tan ^{-1}(0)\right\} = \frac{\pi }{2ab}

therefore \int_{0}^{\pi /2}{\frac{dx}{a^{2}cos^{2}x+b^{2}sin^{2}x}} = \frac{\pi }{2ab} .....................(3)
differentiating (3) wrt a, we get
\int_{0}^{\pi /2}{\frac{-2acos^{2}xdx}{(a^{2}cos^{2}x+b^{2}sin^{2}x)^{2}}} = \frac{-\pi }{2a^{2}b}
\int_{0}^{\pi /2}{\frac{cos^{2}xdx}{(a^{2}cos^{2}x+b^{2}sin^{2}x)^{2}}} = \frac{\pi }{4a^{3}b}.............................(4)
similarly differentiating (3) wrt b we get
\int_{0}^{\pi /2}{\frac{sin^{2}xdx}{(a^{2}cos^{2}x+b^{2}sin^{2}x)^{2}}} = \frac{\pi }{4ab^{3}}.............................(5)
adding (4) and (5) we get
\int_{0}^{\pi /2}{\frac{dx}{(a^{2}cos^{2}x+b^{2}sin^{2}x)^{2}}} = \frac{\pi(a^{2}+b^{2}) }{4a^{3}b^{3}}............................(6)

from (1) and (6) we get

I = \frac{9(2n^{2}+1)\pi .\pi (a^{2}+b^{2})}{n^{2}.4a^{3}b^{3}}
I = \frac{9\pi ^{2}(2n^{2}+1) (a^{2}+b^{2})}{4a^{3}b^{3}n^{2}}

gr8 SR

7) can be done using this initial step of proceeding
\int_{3\pi n}^{3\pi n+3\pi /n}{f(x)}dx=\int_{0}^{3\pi /n}{f(x+3n\pi )}dx where f(x) is given integral

thnks eragon solution to q3 was badly needed and wat a fantastic solution u gave.
i appreciate ur knowledge and THNKS once again [4][4][4]
but can u tell me how to solve Q7) and Q8)

\fn_cm \boxed{Q3}
\fn_cm \\log(1-2acos\theta+a^2)=log\left \{ 1-a\left ( e^{i\theta}+ e^{-i\theta}\right )+a^2 \right \}\\ =log\left \{ \left ( 1-ae^{i\theta} \right )\left ( 1-ae^{-i\theta} \right ) \right \}\\ =log\left ( 1-ae^{i\theta} \right )+log\left ( 1-ae^{-i\theta} \right )\\ =\left ( -ae^{i\theta}-\frac{a^2}{2}e^{2i \theta}-............ \right )+ \left ( -ae^{-i\theta} -\frac{a^2}{2}e^{-2i\theta}-........\right )\\ =-a(e^{i\theta}+e^{-i\theta})-\frac{a^2}{2}(e^{i2\theta}+e^{-i2\theta})-.................\\
\fn_cm =-2acos\theta-\frac{2a^2}{2}cos2\theta-\frac{2a^3}{3}cos3\theta.........................\frac{2a^n}{n}cosn\theta\\ \therefore \\ \int_{0}^{\pi}cosn\theta log\left ( 1-2acos\theta+a^2 \right )d\theta\\ \\=-2\int_{0}^{\pi}\left [ acos\theta cosn\theta+\frac{a^2}{2}cos2\theta cos\ntheta............\frac{a^n}{n}cosn\theta cosn\theta \right ]d\theta\\ =-2\frac{a^n}{n}\int_{0}^{\pi}cos^2n\theta .d\theta
\fn_cm \\\textup{because oder integrals vanish as} \int_{0}^{\pi}cosr\theta cosn \theta .d\theta=0\\\textup{ when } r\neq n

i guess it involves complex nos

anyone Q3) please

the fourth one is very easy:

e^{-\alpha}\left[f(x+\alpha)-f(1+\alpha) \right] \\\\=e^{-\alpha}\int_{1+\alpha}^{x+\alpha}{\frac{e^t}{t}dt}

Now substitute t=z+\alpha

we get what is required.

Actually i said a GAL on goiit not a guy on goiit

but sadly some GUY took it on him....saying that i was reffering to him....

saying saaley ka mooh band kar de ...abusive aint it?

i guess he doesnt know the diff between a guy and a gal....

may god bless him ;)

2) this has been done here and on oder sites many many times.

check this http://www.targetiit.com/iit-jee-forum/posts/integrate-2-13927.html

aur waisey kitna integration karogey bhai....?

i mean such amt of integration is injurious for health [3]

btw i hav seen a gal on goiit who is jus mad abt integrals.....lagta hai wahan sey yahaan sabko bimari lag gayi........No offence [1]