Given `I_m=int_1^e(logx)^mdx ,t h e np rov et h a t(I_m)/(1-m)+m I_(m-2)=e`

IfI_n=int_0^1x^n(tan^(-1)x)dx ,t h e np rov et h a t (n+1)I_n+(n-1)I_(n-2)=-1/n+pi/2

Iff(x)=int_1^x(logt)/(1+t+t^2)dxAAxlt=1,t h e np rov et h a tf(x)f(1/x)dot

If I_1=int_0^pixf(sin^3x+cos^2x)dxa n d I_2=int_0^(pi/2)f(sin^3x+cos^2x)dx ,t h e nr e l a t eI_1a n dI_2

int(1-m)/(x(logx)^(m))dx , xgt0

If int_0^ycost^2dt=int_0^(x^2)(sint)/t dx ,t h ep rov et h a t(dy)/(dx)=(2sinx^2)/(xcosy^2)

Ify=int_0^xf(t)sin{k(x-t)dt ,t h e np rov et h a t(dt^2y)/(dx^2)+k^2y=kf(x)dot

If I=int e^(-x) log(e^x+1) dx, then equal

Let f be a positive function. Let I_1=int_(1-k)^k xf([x(1-x)])dx , I_2=int_(1-k)^kf[x(1-x)]dx ,w h e r e2k-1> 0. T h e n(I_1)/(I_2)i s 2 (b) k (c) 1/2 (d) 1

IfI_(m , n)=int_0^(pi/2)sin^m xcos^n xdx , Then show that I_(m , n)=(m-1)/(m+n)I_m-2n(m ,n in N) Hence, prove that I_(m , n)=f(x)={((n-1)(n-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))pi/4 when both m and n are even ((m-1)(m-3)(m-5)(n-1)(n-3)(n-5))/((m+n)(m+n-2)(m+n-4))

If f(x+2a)=f(x-2a),t h e np rov et h a tf(x)i sp e r iod i cdot