Q12 added [1][1]
1) \int \sqrt{\frac{(1-cos\theta )}{cos\theta (1+cos\theta )(2+cos\theta )}}d\theta
ans..................> cosec^{-1}(2cos^{2}\frac{\theta }{2})+c
[ when solved will add new questions here.]
2) 2) \int \frac{x^{2}}{(xsinx+cosx)^{2}}dx
ans........................>\frac{sinx-xcosx}{cosx+xsinx} + c
Hint : split num. and den. as x cosx and x secx now use by parts
Q3) This one is surely a freak
\int_{0}^{\pi }{ln(1-2acos\theta +a^{2})cosn\theta }d\theta ,where ( a^{2}<1)
nothing mentioned abt ' n '
ans--------> -\frac{\pi a^{n}}{n}
Q4) If f(x) = \int_{0}^{x}{\frac{e^{t}}{t}}dt,x>0
then prove that \int_{1}^{x}{\frac{e^{t}}{(t+\alpha) }}dt=e^{-\alpha }[f(x+\alpha )-f(1+\alpha )]
Q5) \int_{0}^{\pi }{\frac{x^{2}sin2xsin\left(\frac{\pi }{2}cosx \right)}{(2x-\pi )}}dx
ans ------>\frac{8}{\pi }
Q6) \int_{0}^{1}{\frac{dx}{(5+2x-2x^{2})(1+e^{2-4x})}}
ans------> \frac{1}{2\sqrt{11}}ln\left(\frac{1+\sqrt{11}}{\sqrt{11}-1} \right)
Q7) \int_{3n\pi }^{(n^{2}+1)\frac{3\pi }{n}}{\frac{4x}{\left[(a^{2}+b^{2})+(a^{2}-b^{2})cos\frac{2nx}{3} \right]^{2}}}dx [ a,b >0 and n \epsilon I ]
ans---->\frac{9\pi ^{2}(2n^{2}+1)(a^{2}+b^{2})}{4a^{3}b^{3}n^{2}}
Q8) Show that \int_{0}^{2\pi }{\frac{dx}{a+bcosx+csinx}} = \frac{2\pi }{\sqrt{a^{2}-b^{2}-c^{2}}}, when, a>\sqrt{b^{2}+c^{2}}>0
Q9) \int \frac{dx}{(sinx+asecx)^{2}},a\epsilon N
Q10) \int_{0}^{\infty }{log\left(x+\frac{1}{x}\right).\frac{dx}{1+x^{2}} } ans...........> \pi log2 solve using by parts consider log wala as first fuct.
Q11) \int cos\left(blog\left(\frac{x}{a} \right) \right)dx
Q12) \int_{0}^{\pi }{\frac{sin^{2}(2n+1)\theta }{sin^{2}\theta }}d\theta...
I know a similar question was asked earilier but can we do that in the similar way that question was done ??
1thnks yaar.
Bohat time laga post karne main.........
soch raha tha pura post nahin karun magar karte karte pura hi kar diya [3][3]
1ans 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.........
1yup i know that but can that method used der can be applied to this question 12 becoz it's slighty different [7]
1for Q 12 see
http://www.targetiit.com/iit-jee-forum/posts/help-asap-15593.html
29\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
1@ carl
putting x=tanθ
we get
\int log\left( \frac{2}{sin2\theta}\right)d\theta
correct me if i am wrong
1\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]
1I =\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}
1SR UR ANSWER IS RONG .
DIFFERENTIATE KARKE DEKH U WILL GET THE RECIPROCAL OF THE INTEGRAL
ur method is rong
that expressin 1+tanx1-tanx = tan(pi/4 + x)
6omi i did in another method by taking denominator as =t and did it
got diff ans
d same as u told
my question to evryone
hw do i judge wich one is correct or apply wich method in sbsti coz i did another way got another ans...??pls help
29Convert 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
1since 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}}
1gr8 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
1\int_{0}^{2\pi}{} \frac{1}{ a + \sqrt{b^2 + c^2}.sin ( x + \tan^{-1} \frac{c}{b} )}
\textbf{let }x + \tan^{-1} \frac{c}{b} = m
\texttt{ DO NOT CHANGE LIMITS NOW }
\int_{0}^{2\pi}\frac{1}{a + \sqrt{b^2 + c^2 }.sin m }
put, tan ( \frac{m}{2} ) = t
\int_{0}^{2\pi}{\frac{\frac{2dt}{1+ t^2}}{a( 1 + t^2 ) + \sqrt{b^2 + c^2 }.2t}}
\int_{0}^{2\pi}{\frac{ 2dt}{a t^2 + \sqrt{b^2 + c^2 }.2t+ a}}
completing square we get
2 .\int_{0}^{2\pi}{\frac{ dt}{a \left[ \left(t + \frac{\sqrt{b^2 + c^2}}{a} \right)^2+ 1- \frac{b^2 + c^2 }{a^2}}} \right]
2 .\int_{0}^{2\pi}\frac{dt}{a\left[\left(\frac{at+ \sqrt{b^2 + c^2}^}{a} \right)^2+ \frac{a^2 - ( b^2 + c^2 ) }{a^2} \right]}
2 a.\int_{0}^{2\pi}\frac{dt}{ ( at+ \sqrt{b^2 + c^2 }) ^2+ ( a^2 -( b^2 + c^2 ) }
put
at + √b2+ c 2 = k
a .dt = dk
interal reduces to
\frac{2}{\sqrt{a^2 - ( b^2 + c^2 ) }}\int_{0}^{2\pi}{\tan^{-1} \frac{k}{\sqrt{a^2 -b^2 -c^2}}}
\frac{2}{\sqrt{a^2 - ( b^2 + c^2 ) }}\int_{0}^{2\pi}{\tan^{-1} \frac{k}{\sqrt{a^2 -b^2 -c^2}}} = \frac{2\pi}{\sqrt{a^2 - ( b^2 + c^2 ) }}
now substitute value of k and t den put d limits
1thnks 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)
21\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
21Q6..jus use the most impo property of DI i.e I= \int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(a+b-x})dx and the add them
then see what happens. ur integral reduces to this 2I=\int_{0}^{1}{\frac{1}{-2x^2+2x+5}}dx
now its done!
Q5) for this also use the same property applied above
1the 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.
1Actually 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 ;)
12) 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]
