If `I(m,n)=int_0^1 t^m(1+t)^n.dt`, then the expression for `I(m,n)` in terms of `I(m+1,n-1)` is:

If a^m a^n =a^(mn) , then express m in terms of n.

If L(m,n)=int_(0)^(1)t^(m)(1+t)^(n),dt , then prove that L(m,n)=(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)

If a^(m^n)= (a^m)^n , then express m in terms of n.

Find the reduction for I_m,n = int (sin x)^m (cos x)^n dx

If I_m=int_1^x(logx)^mdx satisfies the relation (I_m)=k-l I_(m-1) then

If I_(m)=int_(1)^(e )(log_(e)x)^(m)dx , then the value of (I_(m)+mI_(m-1)) is -

IfI(m , n)=int_0^1x^(m-1)(1-x)^(n-1)dx ,(m , n in I ,m ,ngeq0),t h e n (a) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m-n))dx (b) I(m , n)=int_0^oo(x^(m-1))/((1+x)^(m+n))dx (c) I(m , n)=int_0^oo(x^(n-1))/((1+x)^(m+n))dx (d) I(m , n)=int_0^oo(x^n)/((1+x)^(m+n))dx

If 1/2 log_(3)M + 3log_(3)N = 1 , then express M in terms of N.

If m, n in R , then the value of I(m,n)=int_(0)^(1) t^(m)(1+t)^(n)dt is -

Express M in terms of N: 1/2 log_(3) M +log_(3)N=1