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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 xin 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 xfor m - 1of 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 |
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Answer» `I_(4","2)=1/5 (sin^(3)XCOS^(3)X+8 I_(6","2))` |
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