4U can use the formula directly for sec^n (2x) or this may work out
Take 2x = t
Using IBP
sect ∫(sec^2 (t) dt = sec t tan t
∫sec t tan t (tan t ) dt = ∫ p2 - 1 dp
Put sec t = p
sec t tan t dt = dp
Now I = 1/2(sec t tan t - sec3t - sec t )
hello every one in targetiit
taking some hint from PrItIsH Da
i thot of startin an thread for integration
( similar kind of prctc was folowed at mathlinks too)
will post integrals myself and
d person who answers the
integral question shud post an new integral
( if he has time and an integral in hand )!
PlEaSe PoSt OnLy An I nTeGrAl
hoping that all of u will probably co-operate
1. ∫sec3( 2x ) dx 2.
2.
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11 Answers
442 ) I = ∫ (pi-x) /(a2 - cos2 x ) dx
2I = pi ∫1/(a2 - cos2 x ) dx
Put tanx = p
sec2 x dx = dp
∫ 1/[(a2 - 1) + a2 p2] dp
= 1/(a2 - 1) ∫ 1/[1+kp2] dp ---> I took k = const = a2/(a2-1)
= 1/(a2 - 1) tan-1 (kp) + c
I = (pi/2)[1/(a2 - 1)] tan-1 [(a2)/(a2-1)(tanx)] ] (0 to pi)
= 0
1any one tryin for ques --2
wil post answers if neone doesnt post answer widin today
29SR have asked me many times to post some integrals..so SR i am posting them....
1..\int_{1}^{2n+1}{(x-1)(x-2)(x-3)....{(x-(2n+1))}}dx
where n is a natural number..
2...\int_{0}^{infinity}{\frac{1}{x}sin(x-\frac{1}{x})}dx
3..\int sin^{-1}\frac{2x+2}{\sqrt{4x^{2}+8x+13}} dx
4..\int_{0}^{\pi/4} e^{sinx}(xcosx-secxtanx)dx
5..\int_{0}^{infinity}{\frac{1}{1+x^{2}}ln(x+\frac{1}{x})}dx
6..\int_{0}^{infinity}{\frac{xlnx}{(1+x^{2})^{2}}}dx
For Practise
1\int e^{sinx}xcosx\: dx - \int e^{sinx}secxtanx\: dx
= x\int e^{sinx}cosx\: dx - \int \left(\frac{dx}{dx}\int e^{sinx}cosx\: dx \right)dx-\int e^{sinx}secxtanx\: dx
= xe^{sinx} - \int e^{sinx}dx-\int e^{sinx}secxtanx\: dx
= xe^{sinx} - \int e^{sinx}secxcosx\: dx - \int e^{sinx}secxtanx\: dx
xe^{sinx}-\int secx\: d(e^{sinx})-\int e^{sinx}d(secx)
= xe^{sinx} - \int d(e^{sinx}secx)
= xe^{sinx} -e^{sinx}secx
16)
Integral is : \int_{0}^{\infty }{\frac{xlnx}{(1+x^{2})^{2}}} dx
first i solved it indefinite terms so i took \int{\frac{xlnx}{(1+x^{2})^{2}}} dx
now put 1+x2 = t \Rightarrow 2xdx= dt
now tntegral can be written as \frac{1}{2}\int \frac{2xlnx}{(1+x^{2})^{2}}dx \Rightarrow \frac{1}{2}\int \frac{ln(\sqrt{t-1})}{t^{2}}dt
hence considering \frac{1}{t^{2}} as scond and ln(\sqrt{t-1}) as first function we have integral as
\frac{1}{2}\int \frac{1}{t^{2}}.ln(\sqrt{t-1})dt
now
\Rightarrow \frac{1}{2}\left[ln(\sqrt{t-1}) \int \frac{1}{t^{2}}dt+\int \frac{1}{\sqrt{t-1}}.\frac{1}{2\sqrt{t-1}}.\frac{1}{t}dt\right]
\frac{1}{2}\left[ln(\sqrt{t-1})\left(\frac{-1}{t} \right)+\frac{1}{2}\int \frac{1}{(t-1)t} dt\right]
and this can be finished off now