Let I n,m = 
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 (VI) I n+2,m–2
Suppose we want to establish a relation between I n,m and I n,m – 2 , then we set
P(x) = sin n+1 x cos m–1 x …..(1)
In I n,m and I n,m–2 the exponent of cosx is 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 m – 1 of cosx in P(X). Similarly choose the exponent of sin x for P(x).
Now differentiating both sides of (1), we get
P'(x) = (n + 1) sin n xcos m x –(m–1)sin n+2 xcos m–2 x
= (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 xcos m x
= (n + m) sin n x cos m x – (m – 1) sin n x cos m–2 x
Now integrating both 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.
(i) The relation between I 4,2 and I 2,2 is –
Text Solution
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Ans.
(i)
Sol. Let P = sin 3 x cos 3 x
= 3 sin 2 x cos 4 x – 3 sin 4 x cos 2 x
= 3 sin 2 x (1 – sin 2 x) cos 2 x – 3cos 2 x sin 4 x
= 3sin 2 x cos 2 x – 6sin 4 x cos 2 x
∴ P = 3I 2,2 – 6I 4,2
∴ I 4,2 = 
(– P + 3I 2,2 )
(ii)
Sol. Let P = sins 5 x cos 3 x
∴ 
= 5 sin 4 x cos 4 x – 3sin 6 x cos 2 x
= 5 sin 4 x (1 – sin 2 x) cos 2 x – 3 sin 6 x cos 2 x
= 5 sin 4 x cos 2 x – 8 sin 6 x cos 2 x
∴ P = 5I 4,2 – 8I 6,2
∴ I 4,2 = 
(P + 8I 6, 2 )
(iii)
Sol. Let P = sin 5 xcos 3 x
∴ 
= 5 sin 4 xcos 4 x – 3 sin 6 x cos 2 x
= 5 sin 4 x cos 4 x – 3 sin 4 x(1 – cos 2 x) cos 2 x
= 8 sin 4 x cos 4 x – 3sin 4 x cos 2 x
∴ P = 8I 4,4 – 3I 4,2
∴ I 4,2 = 
(– P + 8I 4,4 )
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