If I n = ∫ d x x 2 + a 2 n where n ∈ I and n > I . If I n and I n – 1 are related by the relation P I n = x x 2 + a 2 n - 1 + Q I n - 1 . Find the value of P and Q in terms of n -

P = 2 a 2 n - 1 and Q = 2 n - 3

P = 2 a 2 n - 1 and Q = 4 n - 3

If I n = ∫ d x x 2 + a 2 n where n ∈ I and n > I . If I n and I n – 1 are related by the relation P I n = x x 2 + a 2 n - 1 + Q I n - 1 . Find the value of P and Q in terms of n -

P = 2 a 2 n - 1 and Q = 2 n - 3

Obtain the reduction formula for ∫ sin n x d x for an integer n ≥ 2 and deduce the value of : ∫ sin 4 x d x

Let I n = ∫ sin n ⁡ x d x = ∫ sin ( n − 1 ) ⁡ x sin ⁡ x d x = ∫ sin n − 1 ⁡ x · d d x ( − cos ⁡ x ) d x = sin ( n − 1 ) ⁡ x ( − cos ⁡ x ) − ∫ ( n − 1 ) sin n − 2 ⁡ x cos ⁡ x ( − cos ⁡ x ) d x = − sin n − 1 ⁡ x cos ⁡ x + ∫ ( n − 1 ) sin n − 2 ⁡ x 1 − sin 2 ⁡ x d x ∵ cos 2 A + sin 2 A = 1 = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) ∫ sin n − 2 ⁡ x d x − ( n − 1 ) ∫ sin n ⁡ x d x = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) I n − 2 − ( n − 1 ) I n = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) I n − 2 − ( n − 1 ) I n Hence, I n = − sin n − 1 ⁡ x cos ⁡ x n + ( n − 1 ) n I n − 2 This is called the reduction formula for ∫ sin n ⁡ x d x If n is even, after successive reduction, we get I 0 = ∫ ( sin ⁡ x ) 0 d x = x + c 1 If n is odd, after successive reduction, we get I 1 = ∫ ( sin ⁡ x ) 1 d x = - cos x + c 2 . Consider, ∫ sin 4 ⁡ x d x On using the reduction formula for ∫ sin n ⁡ x d x with n = 4 and 2 in that order we have, I 4 = ∫ sin 4 ⁡ x d x = − sin 3 ⁡ x cos ⁡ x 4 + 3 4 I 2 = − sin 3 ⁡ x cos ⁡ x 4 + 3 4 − sin ⁡ x cos ⁡ x 2 + 1 2 I 0 = − sin 3 ⁡ x cos ⁡ x 4 − 3 8 sin ⁡ x cos ⁡ x + 3 8 x + c .

Obtain reduction formula for I n = ∫ tan n ⁡ x d x , n being a positive integer n ≥ 2 and deduce the value of ∫ tan 6 ⁡ x d x .

Let I n = ∫ tan n ⁡ x d x = ∫ tan n − 2 ⁡ x tan 2 ⁡ x d x = ∫ tan n − 2 ⁡ x sec 2 x - 1 d x ∵ tan 2 ⁡ x = sec 2 x - 1 = ∫ tan n − 2 ⁡ x sec 2 ⁡ x d x − ∫ tan n − 2 x ⁡ d x = tan n − 1 ⁡ x n − 1 − I n − 2 which is the required reduction formula. When n is even I n will depend on I 0 = ∫ d x = x + c 1 When n is odd I n will depend on I 1 = ∫ tan ⁡ x d x = log ⁡ | sec ⁡ x | + c 2 . Consider, ∫ tan 6 ⁡ x d x I 6 = ∫ tan 6 ⁡ x d x = tan 5 ⁡ x 5 − ∫ tan 4 ⁡ x d x = tan 5 ⁡ x 5 − tan 3 ⁡ x 3 + ∫ tan 2 ⁡ x d x = tan 5 ⁡ x 5 − tan 3 ⁡ x 3 + tan ⁡ x − x + c

If ∫ f ( x ) d x = ψ ( x ) , then ∫ x 5 f ( x 3 ) d x , is equal to

1 3 x 3 ψ ( x 3 ) − ∫ x 2 ψ ( x 3 ) d x + c

1 3 [ x 3 ψ ( x 3 ) − ∫ x 3 ψ ( x 3 ) d x ] + c

1 3 [ x 3 ψ ( x 3 ) − ∫ x 2 ψ ( x 3 ) d x ] + c

1 3 x 3 ψ ( x 3 ) − 3 ∫ x 3 ψ ( x 3 ) d x + c

Obtain reduction formulae for I n = ∫ cot n ⁡ x d x ; n being a positive integer n ≥ 2 and deduce the value of ∫ cot 4 x d x .

Given: I n = ∫ cot n x d x ⇒ I n = ∫ cot n - 2 x · cot 2 x d x ⇒ I n = ∫ cot n - 2 x · cosec 2 x - 1 d x ∵ cosec 2 x - cot 2 x = 1 ⇒ I n = ∫ cot n - 2 x · cosec 2 x d x - ∫ cot n - 2 x d x ⇒ I n = ∫ cot n - 2 x · cosec 2 x d x - I n - 2 Now, put cot x = u ⇒ - cosec 2 x d x = d u {Differentiating w.r.t x } Therefore, I n = - ∫ u n - 2 d u - I n - 2 ⇒ I n = - u n - 2 + 1 n - 2 + 1 - I n - 2 ⇒ I n = - u n - 1 n - 1 - I n - 2 ⇒ I n = - cot n - 1 x n - 1 - I n - 2 (Since u = cot x ) Now, put n = 4 , then I 4 = - cot 3 x 3 - I 2 And again put n = 2 , then I 2 = - cot x - I 0 , where I 0 = ∫ 1 · d x = x Therefore, I 2 = - cot x - x Hence, I 4 = ∫ cot 4 x d x = - cot 3 x 3 - ( - cot x - x ) + C = - cot 3 x 3 + cot x + x + C .

Obtain the reduction formula I n = ∫ sin n ⁡ x d x , n is a positive integer n ≥ 2 and deduce the value of ∫ sin 4 ⁡ x d x .

Let I n = ∫ sin n ⁡ x d x = ∫ sin ( n − 1 ) ⁡ x sin ⁡ x d x = ∫ sin n − 1 ⁡ x · d d x ( − cos ⁡ x ) d x = sin ( n − 1 ) ⁡ x ( − cos ⁡ x ) − ∫ ( n − 1 ) sin n − 2 ⁡ x cos ⁡ x ( − cos ⁡ x ) d x = − sin n − 1 ⁡ x cos ⁡ x + ∫ ( n − 1 ) sin n − 2 ⁡ x 1 − sin 2 ⁡ x d x ∵ cos 2 A + sin 2 A = 1 = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) ∫ sin n − 2 ⁡ x d x − ( n − 1 ) ∫ sin n ⁡ x d x = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) I n − 2 − ( n − 1 ) I n = − sin n − 1 ⁡ x cos ⁡ x + ( n − 1 ) I n − 2 − ( n − 1 ) I n Hence, I n = − sin n − 1 ⁡ x cos ⁡ x n + ( n − 1 ) n I n − 2 This is called the reduction formula for ∫ sin n ⁡ x d x If n is even, after successive reduction, we get I 0 = ∫ ( sin ⁡ x ) 0 d x = x + c 1 If n is odd, after successive reduction, we get I 1 = ∫ ( sin ⁡ x ) 1 d x = - cos x + c 2 . Consider, ∫ sin 4 ⁡ x d x On using the reduction formula for ∫ sin n ⁡ x d x with n = 4 and 2 in that order we have, I 4 = ∫ sin 4 ⁡ x d x = − sin 3 ⁡ x cos ⁡ x 4 + 3 4 I 2 = − sin 3 ⁡ x cos ⁡ x 4 + 3 4 − sin ⁡ x cos ⁡ x 2 + 1 2 I 0 = − sin 3 ⁡ x cos ⁡ x 4 − 3 8 sin ⁡ x cos ⁡ x + 3 8 x + c .