Saturday 2 May 2015

Definite integrals 

0dxx2+a2=π2a
0xp1dx1+x=πsin(pπ), 0<p<1
0xmxn+an=πam+1nnsin[(m+1)π/n], 0<m+1<n
a0dxa2x2=π2
a0a2x2dx=πa24
a0xm(anxn)pdx=am+1+np Γ[(m+1)/n] Γ(p+1)nΓ[(m+1)/n+p+1]
π/20sin2xdx=π/20cos2xdx=π4
0sin(px)xdx=π/2 0π/2p>0p=0p<0
0sin2pxx2=πp2
01cos(px)x2dx=πp2
0cos(px)cos(qx)xdx=lnqp
0cos(px)cos(qx)x2dx=π(qp)2
2π0dxa+bsinx=2πa2b2
2π0dxa+bcos(x)=2πa2b2
0sinax2dx=0cos(ax2)dx=12π2a
0sinxxdx=0cosxxdx=π2
0sin3xx3dx=3π8
0sin4xx4dx=π3
0tanxxdx=π2
π/20dxa+bcosx=arccos(b/a)a2b2

Advanced formulas

π0sin(mx)sin(nx)dx={0π/2m,n integers and mnm,n integers and m=n
π0cos(mx)cos(nx)dx={0π/2m,n integers and mnm,n integers and m=n
π0sin(mx)cos(nx)dx={02m/(m2n2)m,n integers and m+n oddm,n integers and m+n even
π/20sin2mxdx=π/20cos2mxdx=1352m12462mπ2
π/20sin2m+1xdx=π/20cos2m+1xdx=2462m1352m+1
π0sin2p1xcos2q1xdx=Γ(p)Γq2Γ(p+q)
0sin(px)cos(qx)xdx= 0π/2π/4p>q>00<p<qp=q>0
0sin(px)sin(qx)x2dx={πp/2πq/20<pqpq>0
0cos(mx)x2+a2dx=π2aema
0xsin(mx)x2+a2dx=π2ema
0sin(mx)x(x2+a2)dx=π2a2(1ema)
2π0dx(a+bsinx)2=2π0dx(a+bcosx)2=2πa(a2b2)3/2
2π0dx12acosx+a2=2π1a2,  0<a<1
π0xsinxdx12acosx+a2=πaln(1+a)πln(1+1a)|a|<1|a|>1
π0cos(mx)dx12acosx+a2=πam1a2,  a2<1
0sin(axn)dx=1na1/nΓ(1/n)sinπ2n,  n>1
0cos(axn)dx=1na1/nΓ(1/n)cosπ2n,  n>1
0sinxxpdx=π2Γ(p)sin(pπ/2),  0<p<1
0cosxxpdx=π2Γ(p)cos(pπ/2),  0<p<1
0sin(ax2)cos(2bx)dx=12π2a(cosb2asinb2a)
0cos(ax2)cos(2bx)dx=12π2a(cosb2a+sinb2a)
0dx1+tanmxdx=π4
0eaxcosbxdx=aa2+b2
0eaxsinbxdx=ba2+b2
0eaxsinbxxdx=arctanba
0eaxebxxdx=lnba
0eax2dx=12πa
0eax2cosbxdx=12πaeb24a
e(ax2+bx+c)dx=π2eb24ac4a
0xneaxdx=Γ(n+1)an+1
0xmeax2dx=Γ(m+12)2a(m+1)/2
0e(ax2+b/x2)dx=12πae2ab
0xdxex1=π26
0xn1ex1dx=Γ(n)(11n+12n+13n+)
0xdxex+1=π212
0xn1ex+1dx=Γ(n)(11n12n+13n)
0sinmxe2πx1dx=14cothm212m
0(11+xex)dxx=γ
0ex2exxdx=12γ
0(1ex1exx)dx=γ
0eaxebxxsec(px)dx=12ln(b2+p2a2+p2)
0eaxebxxcsc(px)dx=arctanbparctanap
0eax(1cosx)x2dx=arccotaa2ln(a2+1)
10xm(lnx)ndx=(1)nn!(m+1)n+1,m>1,n=0,1,2,
10lnx1+xdx=π212
10lnx1xdx=π26
10ln(1+x)xdx=π212
10ln(1x)xdx=π26
10lnxln(1+x)dx=22ln2π212
10lnxln(1x)dx=2π26
0xp1lnx1+xdx=π2csc(pπ)cot(pπ),0<p<1
10xmxnlnxdx=lnm+1n+1
0exlnxdx=γ
0ex2lnxdx=π4(γ+2ln2)
0ln(ex+1ex1)dx=π24
π/20ln(sinx)dx=π/20ln(cosx)dx=π2ln2
π/20(ln(sinx))2dx=π/20(ln(cosx))2dx=π2(ln2)2+π324
π0xln(sinx)dx=π22ln2
π/20sinxln(sinx)dx=ln21
2π0ln(a+bsinx)dx=2π0ln(a+bcosx)dx=2πln(a+a2b2)
π0ln(a+bcosx)dx=πln(a+a2b22)
π0ln(a22abcosx+b2)dx={2πlna2πlnbab>0ba>0
π/40ln(1+tanx)dx=π8ln2
π20secxln(1+bcosx1+acosx)dx=12(arccos2aarccos2b)

 

No comments:

Post a Comment