" (iii) "(sin^(-1)x)^(m)*(cos^(+)x)^(n),0

If m, n in N , then int_(0)^(pi//2)((sin^(m)x)^(1/n))/((sin^(m)x)^(1/n)+(cos^(m)x)^(1/n))dx is equal to

If m, n in N , then int_(0)^(pi//2)((sin^(m)x)^(1/n))/((sin^(m)x)^(1/n)+(cos^(m)x)^(1/n))dx is equal to

sin^(m)x.cos^(n)x

int_(0)^(pi//2)(sin^(n)x)/((sin^(n)x+cos^(n)x))dx=?

If f(x)={{:(sin(cos^(-1)x)+cos(sin^(-1)x)",",xle0),(sin(cos^(-1)x)-cos(sin^(-1)x)",",xgt0):} then at x = 0

Prove : int sin mx sin n x dx[ m^(2) != n^(2)] , = 1/2 [ (sin(m-n)x)/(m-n) - (sin (m+n)x)/(m+n) ] + c

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