Indefinite Integration FOR IIT-JEE
May 4, 2017 | Author: Apex Institute | Category: N/A
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INDEFINITE INTEGRATION L E V E L - 1 (Objective)
1.
If f(x) =
∫
2 sin x − sin 2 x
(A) 0 2.
If
∫
(B) 1 1 + sin
(A) 2 2 3.
4.
5.
If y =
∫
(
(D) 4 2
)
(A) A = - 1/4 & B may have any value
(B) A = - 1/8 & B may have any value
(C) A = - 1/2 & B = - 1/4
(D) none of these
∫
dx x = I tan-1 m tan + C then : 5 + 4 cos x 2
(B) m = 3
Given (a > 0) ,
∫
∫
(D) m = 2/3
(B) x > e
( ) dx is equal to :
(C) all x ∈ R
(D) no real x .
cot −1 e x e
(C)
∫
(C) I = 1/3
1 dx = loge a loge (loge x) is true for : x log a x
x
( ) +x+c
cot −1 e x 1 2x (A) ln (e + 1) 2 ex −1 x cot e 1
8.
(D) not defined
3/ 2 and y = 0 when x = 0, then value of y when x = 1 is : 1 + x2 1 2 (A) (B) 2 (C) 3 2 (D) 3 2 cos 4 x + 1 If ∫ dx = A cos 4x + B where A & B are constants, then : cot x − tan x
(A) x > 1 7.
(C) 2
x x π dx = A sin − then value of A is : 2 4 4 1 (B) 2 (C) 2 dx
(A) I = 2/3 6.
where x ≠ 0 then Limit f ′ (x) has the value ; x→0
x3
( ) -x+c
2x
2
tan tan
ln (e + 1) -
−1 −1
e
x − cot
−1
x + cot
−1
x x
x
( )
cot −1 e x 1 2x (B) ln (e + 1) + +x+c 2 ex
(D)
1 2
2x
ln (e + 1) +
( ) -x+c
cot −1 e x e
x
dx is equal to :
(A)
4 2 x tan-1 x + ln (1 + x2) - x + c π π
(B)
4 2 x tan-1 x ln (1 + x2) + x + c π π
(C)
4 2 x tan-1 x + ln (1 + x2) + x + c π π
(D)
4 2 x tan -1 x ln (1 + x 2) - x + c π π
9.
If
∫
x4 + 1
(
dx = A ln |x| +
)
x x2 + 1
2
(A) A = 1 ; B = - 1 10.
ln | x|
∫
12.
(B) A = - 1 ; B = 1
1 1 + ln x (ln | x | - 2) + c 3 sin 2 x
2 arctan 2
(C) x -
2 arctan
∫
(D) A = - 1 ; B = - 1
(
(
(B)
2 1 + ln x (ln | x | + 2) + c 3
(D) 2 1 + ln x (3 ln | x | - 2) + c
w.r.t. x is :
1 + sin 2 x
(A) x -
∫
(C) A = 1 ; B = 1
dx equals :
Antiderivative of
)
2 tan x + c
(B) x -
)
(D) x -
2 tan x + c
tan x +c 2 2 tan x +c 2 arctan 2
1
arctan
sin x . cos x . cos 2x . cos 4x . cos 8x . cos 16 x dx equals :
(A) 13.
+ c , where c is the constant of integration then 1 + x2
x 1 + l n |x| 2 1 + ln x (ln | x | - 2) + c (A) 3
(C) 11.
B
sin 16 x +c 1024
(B) -
cos 32 x +c 1024
(C)
cos 32 x +c 1096
(D) -
cos 32 x +c 1096
x 2 + cos 2 x cosec2 x dx is equal to : 2 1+ x
(A) cot x - cot -1 x + c cos ec x
(C) - tan -1 x - sec x
(B) c - cot x + cot -1 x +c
(D) - e ln tan
−1 x
- cot x + c
where 'c' is constant of integration . 14.
15.
16.
∫
3e x + 5e− x dx = Ax + B ln | 4e2x - 5 | + c then : 4 ex − 5e−x
(A)
A = -1, B = -7/8; C = const. of integration
(B)
A = 1, B = 7/8; C = const. of integration
(C)
A = -1/8, B = 7/8 ; C = const. of integration
(D)
A = -1, B = 7/8 ; C = const. of integration
x −1 1 . dx equals : x + 1 x2 1 x2 − 1 (A) sin -1 + x x 2 x −1 (C) sec -1 x +c x dx
∫
∫
x − x2
(B)
1 x2 − 1 + cos -1 + c x x
(D) tan -1
x2 +1 -
x2 −1 +c x
equals :
(A) 2 sin -1 x + c
(B) sin -1 (2x - 1) + c
(C) c - 2 cos -1 (2x - 1)
(D) cos -1 2 x − x 2 + c
17.
∫
when m, n ∈ N is equal to :
2mx . 3nx dx
) e( (B) +c m ln 2 + n ln 3 m ln 2 + n ln 3 x
2 mx + 3nx (A) +c m ln 2 + n ln 3
(C) 18.
19.
∫
2 mx . 3nx
(
ln 2 m . 3 n
)
+c
dx 3
cos x . sin 2x
2 (tan x)5/2 + 2 5
(C)
2 (tan2 x + 5) 5 dx
∫
m ln 2 + n ln 3
+c
equals :
(A)
If
(m n) . 2x . 3x
(D)
sin 3 x cos 5 x
tan x + c
2 (tan2 x + 5) 5
(B)
2 tan x + c
tan x + c
(D) none
= a cot x + b tan 3 x + c where c is an arbitrary constant of integration then
the values of ‘a’ and ‘b’ are respectively : (A) - 2 & 20.
If
∫
2 3
(B) 2 & -
22.
∫
(C) 2 &
(B) sin x
(D) cos 2x
cos 3 x + cos 5 x dx : sin2 x + sin4 x
(A) sin x - 6 tan-1 (sin x) + c
(B) sin x - 2 sin-1 x + c
(C) sin x - 2 (sin x) -1 - 6 tan-1 (sin x) + c
(D) sin x - 2 (sin x)-1 + 5 tan-1 (sin x) + c
∫
ln (tan x) dx equal : sin x cos x
1 2 ln (cot x) + c 2 1 2 ln (sin x sec x) + c (C) 2
24.
(D) none
(C) cos x
(A)
23.
2 3
eu . sin 2x dx can be found in terms of known functions of x then u can be :
(A) x 21.
2 3
π 4 1 π (A) c - cot 2 x + 2 4 1 (C) (tan 4x - sec 4x) + c 2
∫
(B)
1 2 ln (sec x) + c 2
(D)
1 2 ln (cos x cosec x) + c 2
(B)
1 π tan 2 x − + c 2 4
sec2 2 x − dx equals :
Primitive of (A)
x
(
3 x4 − 1
)
x4 + x + 1
x4 + x + 1
+c
2
(B) -
(D) none w.r.t. x is : x x4 + x + 1
+c
(C)
x +1 x + x +1 4
+c
(D) -
x +1 x + x +1 4
+c
L E V E L - 2 (Subjective)
cos 5x + cos 4x dx 1 − 2 cos 3x
1.
∫
4.
dx ∫ (1 − sincotx)x(sec x + 1)
2. ∫ cos x . ex. x2 dx
7. ∫ tan x . tan 2x . tan 3x dx 10. ∫ 13. ∫ 16. ∫
(
ln cos x + cos 2x 2
sin x dx
(x
2
17.
25.
[
31. ∫
3
(x + 2)
(
ex 2 − x2
∫
28. ∫
dx
(x − 1)
)
(1 − x ) 1 − x
3
x
∫
dx
19. ∫ x + x 2 + 2 dx 22. ∫
11.
14.
)
+ x +1
∫
(a > b)
5 x 4 + 4 x5 5
) dx
8.
2
(a + b cos x)
2
5
cos ec x − cot x . cos ec x + cot x
5. ∫
20. ∫
1/ 4
23. ∫
dx
26.
]
29. ∫
dx
32. ∫
3
(1 + x)
34. Integrate
3
1 f ′ (x) 2
1 + 2 sec x
sin x +
∫
cos 2 x sin x
∫
dx
(
cos x
15.
2
]
x 2 + 1 l n x 2 + 1 − 2 ln x x4
(x
dx 3
)
+ 3 x 2 + 3x + 1 x
2
2 − 3x 2 + 3x
x
dx
x 2 + 2x − 3
dx
1+ x dx 1− x
2 − x − x2 2
3/ 2
( x sin x + cos x )
cot x −
∫
dx
w.r.t. x4 , where f (x) = tan -1x + ln 1+ x - ln 1− x
tan x
2
dx
dx
1 + 3 sin 2 x
18. ∫ ex
)
2
cos x
dx
[ (
x
3 12. ∫ e sin x . x cos x 2− sin x dx
dx
)
x −1 4
∫
9.
sin x sin (2 x + α ) sin x
dx
∫ sin x + sec x
dx 6.
dx
∫ (7 x − 10 − x )
1 − x dx 1+ x x
sec x
sin (x − a ) dx sin (x + a )
∫
3.
(x
2
)
+1
( x + 1) 2
dx
21. ∫ l n (ln x) + dx (ln x) 2 1
(ax 2 − b) dx
24.
∫
27.
∫ (x
2
30.
∫ (x
2
33.
∫
x c2 x 2 − (ax 2 + b) 2 x ln x
)
−1
3/ 2
dx
x+2 + 3x + 3
)
dx x +1
dx ( x − α ) ( x − α ) ( x − β)
L E V E L - 3 (Questions asked from previous Engineering Exams)
1.
Find the indefinite integral
∫ (x)
3x + 1 3 . (x + 1)
2.
Evaluate
∫ (x − 1)
3.
Evaluate
∫
f (x) x3 −1
1/ 3
[
]
l n 1 + ( x )1 / 6 1 + + ( x )1/ 4 ( x )1 / 3 + ( x )1 / 2
dx
dx .
dx ; where f(x) is a polynomial of second degree in x such that
f (0) = f (1) = 3 f (2) = - 3 .
∫ cos 2 θ . ln
4.
Evaluate ,
5.
Evaluate
6.
Integrate ,
∫
7.
Let f (x) =
∫
∫
cos 2 x . sin 4 x cos 4 x . (1 + cos 2 2 x)
(
x3 + 3x + 2
)
Evaluate ,
∫
2
x 2 + 1 (x + 1)
dθ .
dx .
dx .
e x (x - 1) (x - 2) d x then f decreases in the interval :
(A) (- ∞ , 2) 8.
cos θ + sin θ cos θ − sin θ
(B) (- 2, - 1)
dx . 2 4 x + 8 x + 13
sin -1
2x + 2
(C) (1, 2)
(D) (2, ∞ )
ANSWER KEY LEVEL - 1 (Objective questions) 1.
b
13.
b,c,d
2.
d
14.
d
3.
d
15.
c
4.
b
16.
a,b,d
5.
a,b
17.
b,c
6.
b
18.
b
7.
c
19.
a
8.
d
20.
c
9.
c
21.
c
10.
a
22.
a,c,d
11.
a
23.
a,b,c
12.
b
24.
b
ANSWER KEY LEVEL - 2 (Subjective Questions) 1.
− (sin x + 1 x e 2
2.
sin 2x )+ c 2
[(x
2
)
− 1 cos x + ( x − 1) 2 . sin x
] +c
(
)
3.
cos a . arc cos cos x - sin a . ln sin x + sin 2 x − sin 2 a + c
4.
1 1 x x x ln tan + sec² + tan + c 4 2 2 2 2
5.
x 1 sin-1 sec 2 + c
6.
1 ln 2 3
7.
1 1 − ln (sec x) − 2 l n (sec 2x) + 3 ln (sec 3x) + c
8.
-
9.
sin x − x cos x +c x sin x + cos x
cos a
2
2
3 + sin x − cos x + arc tan (sin x + cos x) + c 3 − sin x + cos x
[
]
1 2 ln cot x + cot α + cot x + 2 cot α cot x − 1 + c sin α
((
cos 2x - x - cot x . ln e cos x + cos 2x sin x
10.
)) + c
11.
1 t2 − 2 t + 1 1 1 4 ln (1 + t) - ln (1 + t ) + 2 2 ln 2 - tan-1 t2 + c where t = 4 t + 2t +1 2
12.
esinx (x - secx) + c
13.
14.
−
-
b sin x 2a + 2 a − b (a + b cos x) a − b2
(
2
2
)
1 ln 2 2
(
2 cot x − cot 2 x − 1 2 cot x + cot 2 x − 1
)
3/2
+ ln
arc tan
a−b x . tan a+b 2+c
( cot x − 1 + cot x) + c 2
cot x
2 sin 2 x +c sin x + cos x
15.
tan-1
17.
3 x x − 1 3 tan-1 x ln +c 4 16 x + 1 4 x −1 8
19.
1 3
21.
xln (lnx) -
(
x+
x2 + 2
3/ 2
x lnx
8 (x + 1) 2
+
(
2
- x + x2 + 2
)
1/ 2
+c
ex 1 − x + c
27
arcsec x −
28.
ln
29.
8 3
x2 −1
u +u +1 4
2
18.
ex x + 1 + c
5
x −1
(x
2
)
+1
x2 +1 1 . 2 − 3 ln 1 + 2 3 9x x
22.
4 x −1 3 x + 2
2 1 . cos-1 x + 1 + c 16
24.
ax 2 + b sin −1 +k cx
9 7 x − 10 − x 2
4x
2
+ 3 tan −1
1 + 2u 2 3
+ c where u = 3
1− x 1+ x
)
x +c 3 ( x + 1)
1+ x
+
15 ln 8
1+ x −1 1+ x +1
+c
32.
4 − x + 2 2 2 − x − x2 2 − x − x2 2 − sin −1 2x + 1 + c l n − + 3 x x 4
33.
−2 . α −β
34.
- ln (1 - x4) + c
x −β +c x−α
+c
+c
(
15 x 2 + 5x − 2
+c
2 (7 x − 20)
−1 5 t − 1 1 − sin −1 x − 1 − x 2 + c where t = ln tan t + 2 5 5 t + 1
2 arc tan 3
1/ 4
+c
26.
ln x
| u 2 − 1|
-
20.
1+ x
25.
31.
)
x 2 + 2x − 3
23.
30.
)
(
x+1 +c x + x+1
16.
1+ x 1− x
ANSWER KEY LEVEL - 3 (Questions asked from previous Engineering Exams) 1.
I = I1 + I2 + c , where ; y8 8 y 7 28 y 6 56 y 5 70 y 4 56 y 3 28 y 2 I 1 = 12 − + − + − + − 8 y + 1 n y 7 6 5 4 3 2 8 1 1 I 2 = 2 e 3z z − − 9 e 2 z z − + 18 e z (z − 1) − 3 z 2 + c 2 3 2
; where z = ln (1 + x1/6)
1 x +1 x +1 ln +c − 4 x − 1 2( x − 1) 2
2.
3.
x2 + x +1 2 2 x + 1 ln x − 1 + 3 arctan 3 + c
4.
(b)
5.
2 ln (1 + cos 2x) +
6.
x 3 1 1 tan -1 x - ln (1 + x) + ln (1 + x2) + +c 1 + x2 2 2 4
7.
C
8.
-
1 cos θ + sin θ 1 (sin 2 θ ) ln ln (sec 2 θ ) + c 2 cos θ − sin θ 2
2 1 + cos 2 x
; where y = x 1/12 + 1
- ln (1 + cos2 2x) + c
2x + 2 1 3 2x + 2 tan −1 − log ( 4 x 2 + 8 x + 13) 3 2 2 3
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