`I_(m,n)=int(cos^mxcosnxdx=(cos^mxsinx)/(m+n)+f(m,n)I_(m-1,n-1)`,

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))

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)

Let I_(n)=int_(0)^(pi//2) cos^(n)x cos nx dx . Then, I_(n):I_(n+1) is equal to

If I_n=intxcos^nxdx,prove that: I_n=1/n cos^(n-1)x(sinx)+(n-1)/(n)I_(n-2)

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) .

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))}

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(2","2) is

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation I_(4","2) and I_(4","4) is

Let I_(m","n)= int sin^(n)x cos^(m)x dx . Then , we can relate I_(n ","m) with each of the following : (i) I_(n-2","m) " " (ii) I_(n+2","m) (iii) I_(n","m-2) " " (iv) I_(n","m+2) (v) I_(n-2","m+2)" " I_(n+2","m-2) Suppose we want to establish a relation between I_(n","m) and I_(n","m-2) , then we get P(x)=sin^(n+1)x cos^(m-1)x ...(i) In I_(n","m) and I_(n","m-2) the exponent of cos x in m and m-2 respectively, the minimum of the two is m - 2, adding 1 to the minimum we get m-2+1=m-1 . Now, choose the exponent of sin x for m - 1 of cos x in P(x). Similarly, choose the exponent of sin x for P(x)=(nH)sin^(n)x cos^(m)x-(m-1)sin^(n+2) x cos^(m-2)x . Now, differentiating both the sides of Eq. (i), we get =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x(1-cos^(2)x)cos^(m-2)x =(n+1)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x+(m-1)sin^(n)x cos^(n)x =(n+m)sin^(n)x cos^(m)x-(m-1)sin^(n)x cos^(m-2)x Now, integrating both the sides, we get sin^(n+1)x cos^(m-1)x=(n+m)I_(n","m)-(m-1)I_(n","m-2) Similarly, we can establish the other relations. The relation between I_(4","2) and I_(6","2) is