If `I_(m,n)= int(sinx)^(m)(cosx)^(n)`dx then prove that
`I_(m,n) = ((sinx)^(m+1)(cosx)^(n-1))/(m+n) +(n-1)/(m+1). I_(m,n-2)`

If I_(m,n)= int(sinx)^(m)(cosx)^(n) dx then prove that I_(m,n) = ((sinx)^(m+1)(cosx)^(n-1))/(m+n) +(n-1)/(m+n). I_(m,n-2)

If I(m,n)=int_0^1x^(m-1)(1-x)^(n-1)dx , then

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 I_(m,n)= int_(0)^(1) x^(m) (ln x)^(n) dx then I_(m,n) is also equal to

If I_(n)=int cos^(n)x dx . Prove that I_(n)=(1)/(n)(cos^(n-1)x sinx)+((n-1)/(n))I_(n-2) .

If y=(x-a)^(m)(x-b)^(n) , prove that (dy)/(dx)=(x-a)^(m-1)(x-b)^(n-1)[(m+n)x-(an+bm) ].

If I_(n)=int(sinx+cosx)^(n) dx, snd I_(n)=1/n(sinx+cosx)^(n-1)(sinx-cosx)+(2k)/(n) I_(n-2) then k=

The formula in which a certain integral involving some parameters in connected with some integrals of lower order is called a reduction formula. In most of the cases the reduction formula is obtained by the process of integrating by parts. Of course, in some cases the methods of differentiation are adopted.Now answer the question:If I_(m-2,n+2)=intsin^(m-2)xcos^(n+2)xdx and I_(m,n)=-(sin^(m-1)xcos^(n+1)x)/(n+1)+f(m,n)I_((m-2),(n+2)) , then f(2,3) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1/5

If I_(m"," n)=int cos^(m)x*cos nx dx , show that (m+n)I_(m","n)=cos^(m)x*sin nx+m I_((m-1","n-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-2,n)(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))}