Chapter 7 – Integrals

Integrate  – 2 + 3c 3 x 2  w.r.t. x  x x   2a b  Solution ∫  – 2 + 3c 3 x 2  dx  x x  –1 = ∫ 2a ( x ) 2 dx – ∫ bx –2 dx + ∫ 3c x 3 dx = 4a x + b + 9 cx + C . x 5 INTEGRALS 147 3ax

Evaluate ∫ dx b + c2 x2 Solution Let v = b2 + c2x2 , then dv = 2c2 xdx 3ax Therefore, ∫ dx = 3a2 dv b + c2 x2 2c ∫ v 3a = log b 2 + c 2 x 2 + C . 2c 2

Verify the following using the concept of integration as an antiderivative. x 3 dx x 2 x3 ∫ x +1 = x – + – log x + 1 + C d  x 2 x3  Solution  x – + – log x + 1 + C dx  2 3 

Evaluate ∫ dx , x ≠ 1. 1– x 1 x dx Solution Let I = ∫ 1 + x dx = ∫ 2 dx + ∫ = sin –1 x + I1 , 1– x 1– x 1 – x 148 MATHEMATICS where I1 = . 1 – x2 Put 1 – x2 = t2 ⇒ –2x dx = 2t dt. Therefore I1 = – dt = – t + C = – 1 – x 2 + C Hence I = sin–1x – 1 – x 2 + C .

Evaluate ∫ x – α β – x , β > α ( )( ) ( ) Solution Put x – α = t2. Then β – x = β – t 2 + α = β – t 2 – α = – t 2 – α + β and dx = 2tdt. Now 2t dt 2 dt I =∫ =∫ t2 (β – α – t2 ) (β – α – t ) =2 , where k 2 = β – α k – t2 –1 t –1 x–α = 2sin k + C = 2sin β –α + C.

Evaluate ∫ tan x sec x dx 8 4 Solution I = ∫ tan x sec x dx 8 4 ( = ∫ tan x sec x sec x dx ) 2 ( = ∫ tan x tan x + 1 sec x dx ) 2 INTEGRALS 149 = ∫ tan x sec x dx + ∫ tan x sec x dx 10 2 8 2 tan11 x tan 9 x = + +C. 11 9 x3

Find ∫ dx x 4 + 3x 2 + 2 Solution Put x2 = t. Then 2x dx = dt. x3 dx 1 t dt Now I= ∫ = ∫ 2 x + 3x + 2 2 t + 3t + 2

Find ∫ 2sin x + 5cos 2 x Solution Dividing numerator and denominator by cos2x, we have sec 2 x dx I= ∫ 2tan 2 x + 5 150 MATHEMATICS Put tanx = t so that sec2x dx = dt. Then dt 1 dt I = ∫ 2t 2 + 5 = 2 ∫ 2  5 t +   2 1 2  2t  = tan –1   + C 2 5  5 1  2 tan x  = tan –1   + C. 10  5 

Evaluate ∫ ( 7 x – 5) dx as a limit of sums. –1 2 +1 Solution Here a = –1 , b = 2, and h = , i.e, nh = 3 and f (x) = 7x – 5. Now, we have ∫ ( 7 x – 5) dx = lim h  f ( –1) + f (–1 + h) + f ( –1 + 2h ) + ... + f ( –1 + ( n – 1) h ) –1 h →0 Note that f (–1) = –7 – 5 = –12 f (–1 + h) = –7 + 7h – 5 = –12 + 7h f (–1 + (n –1) h) = 7 (n – 1) h – 12. Therefore, ∫ (7x –5) dx = lim h ( –12) + (7h – 12) + (14h –12) + ... + (7 ( n –1 ) h –12) . –1 h→0 = lim h  7 h 1 + 2 + ... + ( n – 1)  – 12n  h→0  INTEGRALS 151  ( n – 1) n  7  = lim h 7 h – .12n  = lim  ( nh )( nh – h ) – 12nh  h →0  2  h → 0 2 

Evaluate ∫ cot 7 x + tan 7 x dx Solution We have π tan 7 x I= ∫ dx ...(1) 0 cot x + tan x 7 7 π π  2 tan 7  – x  2  = ∫ dx by (P4)  π  7π  0 cot 7  – x  + tan  – x  2  2  π cot 7 ( x ) dx =∫ ...(2) cot 7 x dx + tan 7 x Adding (1) and (2), we get π  tan 7 x + cot 7 x  2I = ∫   dx 0 tan 7 x + cot 7 x  π 2 π = ∫ dx which gives I = . 152 MATHEMATICS 10 – x

Find ∫ dx 2 x + 10 – x Solution We have

Find ∫ 1 + sin 2x dx Solution We have π π 4 4 I = ∫ 1 + sin 2 x dx = ∫ ( sin x + cos x )2 dx 0 0 π = ∫ ( sin x + cos x ) dx INTEGRALS 153 π = ( − cos x + sin x ) 4 I = 1.

Find ∫ x tan x dx . 2 –1 Solution I = ∫ x tan x dx 2 –1 1 x3 = tan –1 x ∫ x 2 dx – ∫ . dx 1+ x 3 x3 1  x  = 3 tan x – 3 ∫  x − 1 + x 2  dx –1   x3 x2 1 = tan –1 x – + log 1 + x 2 + C . 3 6 6

Find ∫ 10 – 4 x + 4 x 2 dx Solution We have I = ∫ 10 – 4 x + 4 x 2 dx =∫ ( 2 x – 1) 2 + ( 3) 2 dx Put t = 2x – 1, then dt = 2dx. t 2 + ( 3) dt 2∫ Therefore, I= 1 t2 + 9 9 = t + log t + t 2 + 9 + C 2 2 4 1 9 ( 2 x – 1) ( 2 x – 1) + 9 + log ( 2 x – 1) + ( 2 x – 1) 2 + 9 + C . = 4 4 154 MATHEMATICS

Evaluate ∫ x4 + x2 − 2 . Solution Let x2 = t. Then x2 t t A B 4 2 = 2 = = + x + x − 2 t + t − 2 (t + 2) (t −1) t + 2 t −1 So t = A (t – 1) + B (t + 2) 2 1 Comparing coefficients, we get A = , B= . 3 3 x2 2 1 1 1 So = + x + x − 2 3 x + 2 3 x 2 −1 4 2 2 Therefore, x2 2 1 1 dx ∫ x 4 + x 2 − 2 dx = 3 ∫ x 2 + 2 dx + 3 ∫ x 2 −1 2 1 x 1 x −1 = 3 tan –1 + log +C 2 2 6 x +1 x3 + x Example16 Evaluate ∫x 4 Solution We have x3 + x x3 x dx I= ∫ dx = ∫ dx + 4 x 9 4 x 9 x 9 = I 1+ I 2 . INTEGRALS 155 x3 Now I1 = ∫ 4 x –9 Put t = x4 – 9 so that 4x3 dx = dt. Therefore 1 dt 1 1 I1 = ∫ 4 t = log t + C1 = log x 4 – 9 + C1 4 4 Again, I2 = ∫ x4 – 9 . Put x2 = u so that 2x dx = du. Then 1 du I2 = 2 ∫ u 2 – 3 2 = 2 × 6 log u + 3 + C2 1 u –3 ( ) 1 x2 – 3 = log 2 + C2 .

Show that ∫ = log ( 2 + 1) sin x + cos x 2 Solution We have sin 2 x I= ∫ dx sin x + cos x 156 MATHEMATICS sin 2  – x  π 2  ∫ dx sin  – x  + cos  – x  = 0 π π (by P4) 2  2  π cos 2 x ⇒ I= ∫ dx sin x + cos x π 1 dx Thus, we get 2I = ∫ 2 0 cos  x – π     4 π 1 2  π 1    π  π   2 = ∫  4 sec  x –  dx =  log 2   sec   x –  4 + tan   x –   4  0 1   π π   π  π  =  log  sec + tan  – log sec  –  + tan  −   2  4 4   4  4  2 +1  log ( 2 + 1) – log ( 2 −1)  1 1 2  = = 2 2 –1 1  ( 2 + 1)2  2 = log   = log ( 2 + 1) 2  1  2 Hence I= log ( 2 + 1) . Find ∫ x ( tan –1 x ) dx

INTEGRALS 157 I = ∫ x ( tan –1 x ) dx . Solution Integrating by parts, we have x2  1 tan –1 x tan x  0 – 2 ∫ 2 1 I= ( –1 )  x 2 .2 dx 2  0 1 + x2 π2 x2 = – ∫ 32 0 1 + x 2 .tan –1 x dx π2 x2 = – I1 , where I1 = ∫ tan –1 xdx 32 0 1 + x 2 x2 + 1 – 1 Now I1 = ∫ tan–1x dx 1 + x2 1 1 = ∫ tan x dx – ∫ –1 tan –1 x dx 0 0 1 + x 2 = I2 – ( ( tan –1 x ) )0 1 2 1 π2 = I2 – 2 32 1 1 I2 = ∫ tan x dx = ( x tan –1 x )0 – ∫ –1 1 Here dx 0 0 1 + x2 π 1 ( π 1 ) = – log 1+ x 2 = – log 2 . 4 2 0 4 2 π 1 π2 Thus I1 = – log 2 − 4 2 32 158 MATHEMATICS π2 π 1 π2 π2 π 1 Therefore, I = – + log 2 + = – + log 2 32 4 2 32 16 4 2 π 2 – 4π = + log 2 .

Evaluate ∫ f ( x) dx , where f (x) = |x + 1| + |x| + |x – 1|. –1  2 – x, if –1 < x ≤ 0  Solution We can redefine f as f ( x ) =  x + 2, if 0 < x ≤1  3 x , if 1< x ≤ 2  2 0 1 2 Therefore, ∫ f ( x ) dx = ∫ ( 2 – x ) dx + ∫ ( x + 2 ) dx + ∫ 3x dx –1 –1 0 1 (by P2) 0 1 2  x2   x2   3x 2  =  2x –  +  + 2x  +    2  –1  2 0  2 1  1 1  4 1 5 5 9 19 = 0 –  –2 –  +  + 2  + 3  –  = + + = .  2 2  2 2 2 2 2 2

∫ e ( cos x – sin x ) dx is equal to

• (A)e x cos x + C
• (B)e x sin x + C
• (C)– e x cos x + C
• (D)– e x sin x + C INTEGRALS 159 Solution (A) is the correct answer since ∫ e  f ( x ) + f '( x )  dx = e f ( x ) + C . Here x x f (x) = cosx, f (x) = – sin x.

∫ sin x cos x is equal to 2 2 (A) tanx + cotx + C ( B ) ( t a n x + cotx)2 + C (C) tanx – cotx + C (D) (tanx – cotx)2 + C Solution (C) is the correct answer, since dx ( sin 2 x + cos2 x ) dx I= ∫ = ∫ sin 2 x cos2 x sin 2 x cos 2 x = ∫ sec x dx + ∫ cosec x dx = tanx – cotx + C 2 2 3ex – 5 e– x

If ∫ x dx = ax + b log |4ex + 5e–x| + C, then 4 e + 5 e– x –1 7 1 7

• (A)a = , b=
• (B)a = , b = 8 8 8 8 –1 –7 1 –7
• (C)a = , b=
• (D)a = , b = 8 8 8 8 Solution (C) is the correct answer, since differentiating both sides, we have 3ex – 5 e– x = a + b ( 4 ex – 5 e– x ) , 4 ex + 5 e– x 4 ex + 5 e– x giving 3ex – 5e–x = a (4ex + 5e–x) + b (4ex – 5e–x). Comparing coefficients on both –1 7 sides, we get 3 = 4a + 4b and –5 = 5a – 5b. This verifies a = , b= . 8 8 160 MATHEMATICS b+c

∫ f ( x ) dx is equal to a+c b b

• (A)∫ f ( x – c ) dx
• (B)∫ f ( x + c ) dx a a b b–c
• (C)∫ f ( x ) dx
• (D)∫ f ( x ) dx a a–c Solution (B) is the correct answer, since by putting x = t + c, we get b b I = ∫ f ( c + t ) dt = ∫ f ( x + c ) dx . a a

If f and g are continuous functions in [0, 1] satisfying f (x) = f (a – x) and g (x) + g (a – x) = a, then ∫ f ( x ) . g ( x ) dx is equal to a a

• (A)
• (B)∫ f ( x ) dx 2 0 a a
• (C)∫ f ( x ) dx
• (D)a ∫ f ( x ) dx 0 0 Solution B is the correct answer. Since I = ∫ f ( x ) . g ( x ) dx a a = ∫ f ( a – x ) g ( a – x ) dx = ∫ f ( x ) ( a – g ( x ) ) dx 0 0 a a a = a ∫ f ( x ) dx – ∫ f ( x ) . g ( x ) dx = a ∫ f ( x ) dx – I 0 0 0 INTEGRALS 161 or I = ∫ f ( x ) dx . dt d2y

If x = ∫ and = ay, then a is equal to 0 1 + 9t 2 dx 2

• (A)3
• (B)6
• (C)9
• (D)1 dt dx 1 Solution (C) is the correct answer, since x = ∫ ⇒ = 0 1 + 9t 2 dy 1 + 9 y2 d2y 18 y dy which gives = 2 . = 9y.. dx 2 2 1+ 9y dx x3 + x +1

∫ dx is equal to –1 x 2 + 2 x +1

• (A)log 2
• (B)2 log 2
• (C)log 2
• (D)4 log 2 x3 + x +1 Solution (B) is the correct answer, since I = ∫ dx –1 x 2 + 2 x +1 x +1 x +1 1 1 x3 = ∫ 2 + ∫ dx = 0 + 2 ∫ dx x + 2 x +1 –1 x + 2 x +1 0 ( x + 1) 2 2 –1 [odd function + even function] x +1 1 1 =2 ∫ dx = 2 ∫ dx 1 = 2 log x + 1 0 = 2 log 2. 0 ( x + 1) x +1 162 MATHEMATICS 1 1 et et

If ∫ dt = a, then ∫ dt is equal to 0 (1 + t ) 1+ t 2 e e e e

• (A)a – 1 +
• (B)a + 1 –
• (C)a – 1 –
• (D)a + 1 + 2 2 2 2 Solution (B) is the correct answer, since I = ∫ dt 1+ t 1 1 1 t et 1 + t 0 ∫0 (1 + t ) = e + 2 dt = a (given) et e Therefore, ∫ (1 + t ) = a – 2 + 1.

∫ x cos πx dx is equal to –2 8 4 2 1

• (A)
• (B)
• (C)
• (D)π π π π 2 2 Solution (A) is the correct answer, since I = ∫ x cos πx dx = 2 ∫ x cos πx dx –2 0  12 3    2 2 = 2  ∫ x cos πx dx + ∫ x cos πx dx + ∫ x cos πx dx  = . 0 1 3  π  2 2  Fill in the blanks in each of the Examples 29 to 32. sin 6 x

∫ cos8 x dx = _______. INTEGRALS 163 tan 7 x Solution +C

∫ f ( x ) dx = 0 if f is an _______ function. –a Solution Odd. 2a a

∫ f ( x ) dx = 2 ∫ f ( x ) dx , if f (2a – x) = _______. 0 0 Solution f (x). π sin n x dx

∫ = _______. sin n x + cos n x π