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# Test: Problems On Integration - 1

## 10 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Problems On Integration - 1

Description
This mock test of Test: Problems On Integration - 1 for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: Problems On Integration - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Problems On Integration - 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Problems On Integration - 1 exercise for a better result in the exam. You can find other Test: Problems On Integration - 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### Evaluate:

Solution:

I = ∫√(x² + 5x)dx

= ∫√(x² + 5x + 25/4 - 25/4)

= ∫√{(x + 5/2)² - (5/2)²}

={1/2(x+5/2)(√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

= {(2x + 5)/4 (√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}

Thus, option D is correct...

QUESTION: 2

### Evaluate:

Solution:

sin2x = 1 - cos2x
∫sinx(sin2x - 3cos2x + 15)dx
Put cos2x = t
∫sinx(1 - cos2x - 3cos2x + 15)dx
=  ∫sinx (16 - 4cos2x)dx
Put t = cosx, differentiate with respect to x, we get
dt/dx = -sinx
= -  ∫ [(16 - 4t2)]1/2dt
= -2 ∫ [(2)2 - (t)2]½
= -2{[(2)2 - (t)2]½ + 2sin-1(t/2)} + c
= - cosx {[4 - (cos)2x]½ - 4sin-1(cosx/2)} + c

QUESTION: 3

### Evaluate:

Solution:

Let's apply the integral substitution,

substitute

Now use the standard integral :

substitute back u=(x-2) and add a constant C to the solution,

QUESTION: 4

Evaluate:

Solution:

(x)½ (a - x)½ dx
=  ∫(ax - x2)½ dx
=  ∫{-(x2 - ax)½} dx
=  ∫{-(x2 - ax + a2/4 - a2/4)½} dx
=  ∫{-(x - a/2)2 - a2/4} dx
=  ∫{(a/2)2 - (x - a/2)2} dx
=  ½(x - a/2) {(a/2)2 - (x - a/2)2} + (a2/4) (½ sin-1(x - a)/a2)
= {(2x - a)/4 (ax - x2)½} + {a2/8 sin-1(2x - a)/a} + c

QUESTION: 5

is equal to

Solution:

QUESTION: 6

Evaluate:

Solution:
QUESTION: 7

Solution:
QUESTION: 8

Evaluate:

Solution:

Apply Trig Substitution: x=4sin(u)​

QUESTION: 9

Evaluate:

Solution:

Let y = ex
dy/dx = ex dx
= ∫(9 - y2)^½ dy
= ∫[(3)2 - (y)2]½ dy
Apply [(a)2 - (x)2] formula
⇒ y/2[(3)2 - (y)2]½ + [(3)2]/2 sin-1(y/3) + c
= ex/2[(3)2 - (y)2]½ + [9]/2 sin-1(ex/3) + c

QUESTION: 10

The integral of the function  w.r.t. x is:

Solution: