Integrals- Quiz 2

Integrals Quiz-2

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Q1. If I=

|sin⁡x| [sin⁡x]dx (where [.] denotes the greatest integer function), then the value of I is

•  -40
•  40
•  20
•  -20

Solution
∫(-20π)^20π|sin⁡x |[sin⁡x ]dx  =∫0^20π|sin⁡x |([sin⁡x]+[–sin⁡x]) dx   =-20∫_0^π(sin⁡x )dx=-20 (–cos⁡x )0^π=20 (-2)=-40

Q2.If I=∫(√(cot⁡x )-√(tan⁡x ))dx,then I equals

•  √2 log⁡(√(tan⁡x )-√(cot⁡x ))+C
•  √2 log⁡|sin⁡x+cos⁡x+√(sin⁡2x )|+C
•  √2 log⁡|sin⁡x-cos⁡x+√2 sin⁡x cos⁡x|+C
•  √2log⁡|sin⁡(x+π/4)+√2 sin⁡x cos⁡x|+C

Solution
I=∫cos⁡x-sin⁡x/√(cos⁡x sin⁡x) dx  Put sin⁡x+cos⁡x=t,so that 2 sin⁡x cos⁡x= t^2 -1  ∴ I= √(2 ) ∫dt/√(t^(2 )- 1) = √(2 ) log⁡|t +√(t^(2 )- 1)|+c  =√2 log⁡|sin⁡x+cos⁡x+√(sin⁡2x ) |+C

Q3.  The value of the definite integral

√(tan⁡x )dx is

•   √2 π
•   π/√2
•  2√2 π
•  π/2√2

Solution

Q4. Let f(x)=x/(1+x^n )^(1⁄n) for n≥2 and g(x)=(fofo…of)(x).Then, ∫x^(n-2)g(x)dxequals

•  1/n(n-1)(1+nx^n )^(1-1/n)+c
•  1/(n-1)(1+nx^n )^(1-1/n)+c
•  1/(n(n+1))(1+nx^n )^(1+1/n)+c
•  1/(n+1)(1+nx^n )^(1+1/n)+c

Solution
Here, ff(x)=(f(x))/[1+f(x)^n ]^(1⁄n) =x/(1+2x^n )^(1⁄n)   and fff(x)=x/(1+3x^n )^(1⁄n) nbsp; ∴ =x/(1+nx^n )^(1⁄n)   Let I=∫x^(n-2) g(x)dx=∫(x^(n-1) dx)/(1+nx^n )^(1⁄n)   =1/n^2 ∫(n^2 x^(n-1) dx)/(1+nx^n )^(1⁄n)  =1/n^2 ∫(d/dx(1+nx^n))/(1+nx^n )^(1⁄n) dx ∴ I=1/n(n-1) (1+nx^n )^(1-1/n)+c

Q5.If λ =

is equal to

•  2λ
•  e loge⁡ 2-λ
•  λ
•  e loge⁡ 2+λ

Solution
 Given λ=∫0^1e^t/(1+t) dt  ∫0^1e^t log_e⁡(1+t)dt=[log_e⁡(1+t) e^t ]0^1-∫0^1e^t/(1+t)=e loge⁡ 2-λ 

Q6. ∫1/√(sin^3 x sin⁡(x+α))dx,α≠nπ,n∈Z is equal to

•  -2 cosec α(cos⁡α-tan⁡x sin⁡α)^(1/2)+C
•  -2 (cos⁡α+cot x sin⁡α )^(1/2)+C
• -2cosecα(cos⁡α+cot⁡x sin⁡α)^(1/2)+C
•  -2cosec α(sin⁡α+cot⁡x cos⁡α)^(1/2)+C

Solution
sin^3 x sin⁡(x+α)  =sin^3 x(sin⁡x cos⁡α+cos⁡x sin⁡α)  =sin^4 x(cos⁡α+cot⁡x sin⁡α )  I=∫1/√(sin^3 x sin⁡(x+α) ) dx  =∫1/(sin^2 x √(cos⁡α+cot⁡x sin⁡α )) dx  =∫(cosec^2 x)/√(cos⁡α+cot⁡x sin⁡α ) dx  Putting cos⁡α+cot⁡x sin⁡α=t and-cosec^2 x sin⁡αdx=dt, we have  I=∫-1/sin⁡α√t dt=-1/sin⁡α ∫t^(-1/2) dt  =1/sin⁡α (t^(1/2)/(1/2))+C  ⇒I=-2 cosec α √(t )+C  =-2 cosec α(cos⁡α+cot⁡x sin⁡α )^(1/2)+C 

Q7.∫sin⁡2x/sin⁡5x sin⁡3x dx is equal to

•  log sin⁡3x-log⁡ sin⁡5x+c
•  1/3 log⁡ sin⁡3x+ 1/(5 )log⁡ sin⁡5x+c
•  1/3 log⁡ sin⁡3x- 1/(5 )log⁡ sin⁡5x+c
•  3 log ⁡sin⁡ 3x-5 log⁡ sin⁡5x+c

Solution
∫sin⁡2x/sin⁡5x sin⁡3x dx  =∫sin⁡(5x-3x)/sin⁡5x sin⁡3x   =∫sin⁡5x cos⁡3x-cos⁡5x sin⁡3x/sin⁡5x sin⁡3x dx  =1/3 log⁡sin⁡3x-1/5 log⁡sin⁡5x+ C 

Q8.∫sin⁡2x/(sin^4⁡x+cos^4⁡x ) dx is equal to

•  cot^(-1) (tan^2⁡x )+c
•  tan^(-1) (tan^2⁡x )+c
•  cot^(-1) (cot^2⁡x )+c
•  tan^(-1) (cot^2⁡x )+c

Solution
I=∫sin⁡2x/sin^4⁡x+cos^4⁡x dx  =∫(2 sin⁡x cos⁡x )/sin^4⁡x+cos^4⁡x dx  =∫(2 tan⁡x sec^2⁡x )/(1+tan^4 x) dx  Let tan^2 x=t⇒2tan⁡x sec^2⁡x dx=dt  ⇒I=∫dt/(1+t^2 )=tan^(-1)⁡+C=tan^(-1)⁡(tan^2⁡x ) +C 

Q9.∫(3+2 cos⁡x)/(2+3 cos⁡x )^(2 ) dx is equal to

•  (sin⁡x/(3 cos⁡x+2 ))+c
•  ((2 cos⁡x)/(3 sin⁡x+2 ))+c
•  ((2 cos⁡x)/(3 cos⁡x+2))+c
•  ((2 sin⁡x)/(3 sin⁡x+2))+c

Solution
Let I=∫(3+2 cos⁡x)/(2+3 cos⁡x)^2 dx, Multiplying N^rand  D^(r )by cosec^2 x, we get  ⇒I=∫((3 cosec^2 x+2 cot⁡x cosec x))/(2 cosec x+cot⁡x)^2 dx  =-∫(-3cosec^2 x-2 cot⁡x cosec x )/(2 cosec x+3 cot⁡x)^2 dx  =1/(2 cosec x+3 cot⁡x )+C=(sin⁡x/(2+3 cos⁡x ))+C

Q10. Let f be a non-negative function defined on the interval [0,1]. If

and f(0)=0, then

•  f(1/2)<1/2 and f(1/3)>1/3
•  f(1/2)>1/2 and f(1/3)>1/3
•  f(1/2)<1/2 and f(1/3)<1/3
• f(1/2)>1/2 and f(1/3)<1/3

Solution
Given, ∫0^x√(1-(f^' (t) )^2 ) dt=∫0^xf(t)dt,0≤x≤1 Applying Leibnitz theorem, we get  √(1-(f^' (x) )^2 )=f(x)  ⇒1-(f^' (x) )^2=f^2 (x)  ⇒ (f^' (x) )^2=1-f^2 (x)  ⇒f^' (x)=±√(1-f^2 (x))  ⇒ dy/dx=±√(1-y^2 ),where y=f(x)  ⇒ dy/√(1-y^2 )=±dx  On integrating both sides, we get  sin^(-1)⁡(y)=±x+C  ∵f(0)=0⇒C=0  ∴y=±sin⁡x  y=sin⁡x=f(x)given f(x)≥0 for x∈[0,1]  It is known that sin⁡x less than x,∀ x∈R^+  ∴sin⁡(1/2)<1/2⇒f(1/2)<1/2 and sin⁡(1/3)<1/3  ⇒ f(1/3)<1/3