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

Evaluate: ∫ 1 cos h x + sin h x d x on R .

Consider, I = ∫ 1 cos h x + sin h x d x ⇒ I = ∫ 1 cos h x + sin h x × cos h x - sin h x cos h x - sin h x d x ⇒ I = ∫ cos h x - sin h x cos h 2 x - sin h 2 x d x ⇒ I = ∫ cos h x - sin h x 1 d x ∵ cos h 2 x - sin h 2 x = 1 ⇒ I = ∫ cos h x d x - ∫ sin h x d x ⇒ I = sin h x - cos h x + c ∴ I = sin h x - cos h x + c , where c is a constant.

Evaluate: ∫ 5 - x x - 2 d x on ( 2 , 5 ) .

Let I = ∫ 5 - x x - 2 d x ⇒ I = ∫ 5 - x x - 2 × 5 - x 5 - x d x ⇒ I = ∫ 5 - x 5 x - x 2 - 10 + 2 x d x = ∫ 5 - x 7 x - x 2 - 10 d x We know that, ∫ p x + q a x 2 + b x + c d x = p x + q = A d d x a x 2 + b x + c + B ∴ ∫ 5 - x 7 x - x 2 - 10 d x = 5 - x = A d d x 7 x - x 2 - 10 + B ⇒ 5 - x = A ( 7 - 2 x ) + B On equating the coefficients and constants, we get A = 1 2 , B = 3 2 ∴ 5 - x = 1 2 ( 7 - 2 x ) + 3 2 ∴ I = ∫ 5 - x 7 x - x 2 + 0 d x = 1 2 ∫ 7 - 2 x 7 x - x 2 - 10 d x + 3 2 ∫ 1 7 x - x 2 - 10 d x We know that ∫ f ' ( x ) f ( x ) d x = 2 f x and consider I 1 = ∫ 1 7 x - x 2 - 10 d x = ∫ d x - x 2 - 7 x + 10 = ∫ d x - x 2 - 7 x + 7 2 2 - 7 2 2 + 10 = ∫ d x - x - 7 2 2 + 3 2 2 ⇒ I 1 = ∫ d x 3 2 2 - x - 7 2 2 = sin - 1 ( x - 7 2 ) 3 2 + c Hence, I = 7 x - x 2 + 10 + 3 2 sin - 1 2 x - 7 3 + 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 .

Evaluate: ∫ cos ⁡ x ( 1 + sin ⁡ x ) 2 d x

Let I = ∫ cos ⁡ x ( 1 + sin ⁡ x ) 2 d x Put 1 + sin ⁡ x = t ⇒ cos ⁡ x d x = d t ∴ I = ∫ d t t 2 = − 1 t + C = − 1 1 + sin ⁡ x + C

Evaluate: ∫ x log ⁡ x d x on ( 0 , ∞ )

Let I = ∫ x log ⁡ x d x Applying integration by parts by taking u ( x ) = log ⁡ x , v ( x ) = x 2 2 . Then, v ' ( x ) = x . Using ILATE, we get: ∫ x log ⁡ x d x = x 2 2 log ⁡ x − ∫ 1 x ⋅ x 2 2 d x = x 2 2 log ⁡ x − 1 2 ⋅ x 2 2 + c = x 2 2 log ⁡ x − 1 4 x 2 + c

Evaluate: ∫ 1 1 + sin ⁡ x + cos ⁡ x d x

Let I = ∫ 1 1 + sin x + cos x d x = ∫ 1 1 + 2 tan x 2 1 + tan 2 x 2 + 1 - tan 2 x 2 1 + tan 2 x 2 d x ∵ sin 2 A = 2 tan A 1 + tan 2 A , cos 2 A = 1 - tan 2 A 1 + tan 2 A = ∫ 1 + tan 2 x 2 1 + tan 2 x 2 + 2 tan x + 1 - tan 2 x 2 d x = ∫ sec 2 x 2 d x 2 + 2 tan x 2 ∵ 1 + tan 2 A = sec 2 A = 1 2 ∫ sec 2 x 2 d x 1 + tan x 2 Put tan x 2 = t ⇒ 1 2 · sec 2 x 2 = d t d x ⇒ d t = 1 2 · sec 2 x 2 d x ∴ I = ∫ d t 1 + t = ln | 1 + t | + C Hence, I = ln 1 + tan x 2 + C

If I n = ∫ cos n x d x , then show that I n = 1 n cos n - 1 x sin x + n - 1 n I n - 2 . (where n ≥ 2 )

Given: I n = ∫ cos n x d x ⇒ I n = ∫ cos n - 1 x cos x d x Using Integration By parts, ∫ u v d x = u ∫ v d x - ∫ u ' ∫ v d x d x Here, u = cos n - 1 x and v = cos x (Using ILATE principle) Therefore, I n = cos n - 1 x ∫ cos x d x - ∫ d d x cos n - 1 x ∫ cos x d x d x ⇒ I n = cos n - 1 x sin x - ∫ sin x ( n - 1 ) cos n - 2 x ( - sin x ) d x ∵ ∫ cos a x d x = sin a x a + C , d d x [ f ( x ) ] n = n [ f ( x ) ] n - 1 f ' ( x ) ⇒ I n = cos n - 1 x sin x + ( n - 1 ) ∫ cos n - 2 x 1 - cos 2 x d x ∵ cos 2 A + sin 2 A = 1 ⇒ I n = cos n - 1 x sin x + ( n - 1 ) ∫ cos n - 2 x d x - ∫ cos n x d x ⇒ I n = cos n - 1 x sin x + ( n - 1 ) I n - 2 - ( n - 1 ) I n ⇒ I n ( 1 + n - 1 ) = cos n - 1 x · sin x + ( n - 1 ) I n - 2 ⇒ I n = cos n - 1 x sin x n + n - 1 n I n - 2 Hence, proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Calculus>Chapter 6 - Integration>Andhra Pradesh Board-2019

Embibe Experts Solutions for Exercise 1: Andhra Pradesh Board-2018

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