1yup i know that but can that method used der can be applied to this question 12 becoz it's slighty different [7]
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 ??
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30 Answers
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.........
11for 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]