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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
` int _(-2)^(2)|x cos pi x |dx` is equal toA. `8/pi`B. `4/pi`C. `2/pi`D. `1/pi` |
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Answer» Correct Answer - A Let `l - int _(-2)^(2) | x cos pi x| dx = 2 = 2 int _(0)^(2) | x cos pix| dx ` `= 2 { int _(0)^(1/2) | x cos pi x | dx + int _(1/2)^(3/2) | x cos pi x | dx + int _(3/2)^(2)| x cos pi x | dx }` ` = 2 [ int _(0)^(1//2) cos pi xdx + int _(1//2)^(3/2) x cospi x dx + int _(3//2)^(2) x cos pi xdx]` ` = 2 [ [ (xsin pix)/pi + (cos pi x)/(pi^(2))]_(0)^(1//2) - [ (x sin pi x)/pi + (cos pi x)/(pi^(2))] _(1//2)^(3//2)` ` + [ (x sin pi x)/pi + (cos pix)/(pi^(2))]_(3//2)^(2)]` ` = 2 [ (1/(2pi)- 1/(pi^(2)))-((-3)/(2pi)-1/(2pi))+(1/(pi^(2))+3/(2pi)) ] ^(1//2) = 2 xx 8/(2pi) = 8/(pi)` |
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| 102. |
Given function `f(x) ={:{(x^(2),"for " 0 lexlt1),(sqrt(x),"for "1 le x le2):}, "then" int_(0)^(2)f(x)dx` is equal toA. ` (4sqrt(2)-1)`B. `1/3(4sqrt(2)-1)`C. ` 1/3 (sqrt(2)-1)`D. None of these |
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Answer» Correct Answer - B Here, ` int _(0)^(pi)f(x) dx = int _(0)^(1) f(x) dx + int _(1)^(2) f(x) dx ` ` :. int _(0)^(2) f(x) dx = int_(0)^(1) x^(2) sqrt(x) dx` ` rArr int _(0)^(2) f(x) dx = [ (x^(3))/3]_(0)^(1)+ [ 2/3 x sqrt(x)]_(1)^(2)` ` = [ 1/3 - 0 ] +2/3 [ 2 sqrt(2)-1]` ` = 1/3 + (4sqrt(2))/2 - 2/3 = (4sqrt(2))/3 -1/3 ` |
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| 103. |
`int_(0)^(1)(dx)/(e^(x)+e^(-x))`A. `(1-(pi)/(4))`B. `tan^(-1)e`C. `tan^(-1)e+(pi)/(4)`D. `tan^(-1)e-(pi)/(4)` |
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Answer» Correct Answer - D `I=int_(0)^(1)(dx)/((e^(x)+(1)/(e^(x))))=int_(0)^(1)(e^(x)dx)/((e^(2x)+1))=int_(0)^(e)(dt)/((t^(2)+1))`, where `e^(x)=t`. `=[tan^(-1)t]_(1)^(e)=(tan^(-1)e-(pi)/(4))`. |
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| 104. |
`int_(0)^(pi//2) x sin x dx ` is equal to |
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Answer» Correct Answer - B ` int _(0)^(pi//2) x sin x dx = [ x(-cos x)] _(0)^(pi//2) - int _(0)^(pi//2) 1 ( - cos x) dx` , ` = 0 + [ sin x] _(0)^(pi//2) =1` |
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| 105. |
` int_(0)^(pi) sqrt((cos2x+1)/2)dx` is equal to |
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Answer» Correct Answer - B ` int _(0)^(pi) sqrt((cos 2x +1)/2) dx = int _(0)^(pi) sqrt(cos^(2)x)dx = int _(0)^(pi) | cos x | dx ` ` = int _(0)^(pi//2) cos x dx - int _(pi//2) ^(pi) cos x dx = [ sin x] _(0)^(pi//2) - [ sin x ] _(pi//2)^(pi) = 2 ` |
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| 106. |
`int_(0)^(pi)(sin2xcos3x)dx=?`A. `(4)/(5)`B. `-(4)/(5)`C. `(5)/(12)`D. `-(12)/(5)` |
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Answer» `2cosAsinB=sin(A+B)-sin(A-B)` `:.I=(1)/(2)int_(0)^(pi)(sin5x-sinx)dx=(1)/(2)*[(-cos5x)/(5)+cosx]_(0)^(pi)=(-4)/(5)`. |
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| 107. |
`int_(1)^(x) (logx^(2))/x dx` is equal toA. `(logx)^(2)`B. `(1)/(2)(logx)^(2)`C. `(logx^(2))/(2)`D. None of these |
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Answer» Correct Answer - A Let ` l = int _(1)^(x) (2log x)/x dx` Put log ` x= t rArr (dx)/x = dt ` ` :. l = 2 int _(0)^(1ogx) t dt = 2 [(t^(2))/2]_(0)^(logx)=(log x)^(2)` |
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| 108. |
` int_(-1)^(1) sin^(3) x cos^(2) x ` dx is equal toA. `-1`B. 1C. 0D. None of these |
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Answer» Correct Answer - C Let ` l = int _(-1)^(1) sin^(3) x cos^(2) x dx` , Again , let f(x) `= sin^(3) x cos^(2) x` ` f(-x) = sin^(3) (-x) cos^(2) (-x) = - sin^(3) x cos^(2) x` , ` :. F(x) ` is an odd function , Hence, ` int_(-1)^(1) sin^(3) x cos^(2) x dx = 0` |
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| 109. |
`int_(-a)^(a)x^(3)sqrt(a^(2)-x^(2))dx=0` |
| Answer» `f(-x)=-f(x)impliesI=0`. | |
| 110. |
`int_(0)^(pi//2)sinxsin2xdx=?`A. `(2)/(3)`B. `(3)/(4)`C. `(5)/(6)`D. `(3)/(5)` |
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Answer» `I=(1)/(2)int_(0)^(pi//2)[cos(2x-x)-cos(2x+x)]dx=(1)/(2)int_(0)^(pi//2)(cosx-cos3x)dx` `=(1)/(2)[sinx-(sin3x)/(3)]_(0)^(pi//2)=(1)/(2)(1+(1)/(3))=(2)/(3)`. |
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| 111. |
`int_(0)^(a)(dx)/(x+sqrt(a^(2)-x^(2)))=(pi)/(4)` |
| Answer» Put `x=a sin theta` and `dx=a cos theta d theta`. | |
| 112. |
The value of `int_(1)^(e^(2)) (dx)/(x(1+logx)^(2))`isA. `(2)/(3)`B. `(1)/(3)`C. `(3)/(2)`D. `ln 2` |
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Answer» Correct Answer - A Let ` l = int_(1)^(e^(2)) (dx)/(1+log x)^(2)` Put ` 1+ log x = t` ` rArr 1/x dx = dt` , ` :. L = int_(1)^(3) 1/(t^(2)) dt = [ -1/t]_(1)^(3) =2/3` |
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| 113. |
`int_0^oo x/((1+x)(1+x^2)) dx` |
| Answer» Putting `x=tan theta`, we get `I=int_(0)^(pi//2)(sin theta)/((costheta+sintheta))d theta`. | |
| 114. |
`int_(-pi)^(pi)(sin^(75)x+x^(125))dx=0` |
| Answer» `f(-x)=-f(x)impliesI=0`. | |
| 115. |
`int_(0)^(a)(sqrt(x))/((sqrt(x)+sqrt(a-x)))dx=(a)/(2)` |
| Answer» Apply `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`. | |
| 116. |
`int_0^(pi//6)cosxcos2x dx`A. `(1)/(4)`B. `(5)/(12)`C. `(1)/(3)`D. `(7)/(12)` |
| Answer» `I=(1)/(2)int_(0)^(pi//6)[cos3x+cosx]dx=(1)/(2)[(sin3x)/(3)+sinx]_(0)^(pi//6)=(15)/(12)`. | |
| 117. |
`int_(0)^(2) (2x-2)/(2x-x^(2)) `dx is equal to |
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Answer» Correct Answer - A Let ` l = int_(0)^(2) (2x-2)/(2x-x^(2)) dx` , Put ` 2x - x^(2) = t rArr (2- 2x) dx = dt` ` l = -int_(0)^(0) (dt)/t = 0` , |
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| 118. |
Evaluate `int_0^1 (1-x)/(1+x) dx`A. `(1)/(2)log2`B. `(2log2+1)`C. `(2log2-1)`D. `((1)/(2)log2-1)` |
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Answer» On dividing `(-x+1)` by `(x+1)` we get : `I=int_(0)^(1){-1+(2)/((x+1))}dx=[-x+2log(x+1)]_(0)^(1)=(2log2-1)`. |
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| 119. |
सिद्ध कीजिए की `int_(0)^(pi) (x dx)/(1+ sin^(2) x) = pi^(2)/(2sqrt(2))`. |
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Answer» `2I=2pi*int_(0)^(pi//2)(dx)/((1+sin^(2)x))`. Divide num. and denom. by `cos^(2)x` and put `tanx=t`. |
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| 120. |
`int_(0)^(pi//4) (sec^(2)x)/((1+tan x)(2+tan x))dx` is equal toA. `log_(e)((2)/(3))`B. `log_(e)3`C. `(1)/(3)log_(e)((4)/(3))`D. `log_(e)((4)/(3))` |
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Answer» Correct Answer - D Put `1+tanx=t rArr sec^(2)xdx=dt` `therefore" "int_(0)^(pi//4)(sec^(2)x)/((1+tanx)(2+tanx))dx` `=int_(1)^(2)(dt)/(t(1+t))=int_(1)^(2)(dt)/(t)-int_(1)^(2)(dt)/(1+t)` `=[logt-log(1+t)]_(1)^(2)` `=log_(2)2-log_(e)3+log_(e)2=log_(e).(4)/(3)`. |
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| 121. |
`int_(-pi)^(pi)x^(12)sin^(9)xdx=0` |
| Answer» `f(-x)=-f(x)impliesI=0`. | |
| 122. |
`int_(0)^(pi//2)sin^(2)xdx=?`A. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(2)`D. `(2pi)/(3)` |
| Answer» `I=(1)/(2)*int_(0)^(pi//2)(1-cos2x)dx=(1)/(2)[x-(sin2x)/(2)]_(0)^(pi//2)=(pi)/(4)`. | |
| 123. |
`int_(0)^(pi)sin^(2)xcos^(3)xdx=0` |
| Answer» Apply `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` to get `I=-I`. | |
| 124. |
`int_(1)^(x) (log(x^(2)))/(x)` dx is equal toA. `(logx)^(2)`B. `(1)/(2)(logx)^(2)`C. `(logx^(2))/(2)`D. None of these |
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Answer» Correct Answer - A Let ` l = int_(1)^(x) (log(x^(2)))/x dx = 2 int _(1)^(x) log x.1/x dx` , Put ` log x = t rArr 1/x dx = dt ` , ` :. l = 2 int _(0)^(log x) t dt = 2 [ (t^(2))/2]_(0)^(logx) = (logx)^(2)` |
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| 125. |
`int_(0)^(pi)(x sinx)/((1+sinx))dx=pi((pi)/(2)-1)` |
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Answer» `2I=pi*int_(0)^(pi)(sinx)/((1+cosx))dx=pi*int_(0)^(pi)20(1-(1)/(1+sinx))dx` `=pi^(2)-pi*int_(0)^(pi)((1-sinx)/(1-sin^(2)x))dx=pi^(2)-pi*int_(0)^(pi)((1-sinx))/(cos^(2)x)dx` `pi^(2)-pi*int_(0)^(pi)(sec^(2)x-secxtanx)dx`. |
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| 126. |
`int_(-1)^(0) (dx)/(x^(2) + 2x + 2 )` is equal to |
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Answer» Correct Answer - B ` int_(-1)^(0) (dx)/(x^(2)+2x + 2 ) = int _(-1)^(0) (dx)/((x+1)^(2)+1)` ` = [ tan^(-1) (x+1)] _(-1)^(0) = [ tan ^(-1) 1 - tan ^(-1) 0 ] = pi/4 ` |
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| 127. |
`int_(0)^(pi//2)(sinxcosx)/((1+sin^(4)x))dx` |
| Answer» Correct Answer - Put `sin^(2)x=t`. | |
| 128. |
`int_(-a)^(a)log((a-x)/(a+x))dx=?`A. `2a`B. `a`C. `0`D. `1` |
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Answer» `f(x)=log((a-x)/(a+x))`. `impliesf(-x)=log((a+x)/(a-x))=log{((a-x)/(a+x))^(-1)}=-log((a-x)/(a+x))=-f(x)` `:.f(x)` is odd and hence `I=0`. |
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| 129. |
`int_(-1)^(1)e^(|x|)dx=2(e-1)` |
| Answer» `I=int_(-1)^(0)e^(-x)dx+int_(0)^(1)e^(x)dx`. | |
| 130. |
`int_(0)^(pi)(x tanx)/((secx+cosx))dx=(pi^(2))/(4)` |
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Answer» `I=int_(0)^(pi)(xsinx)/((1+cos^(2)x))dx` and `I=int_(0)^(pi)((pi-x)sinx)/((1+cos^(2)x))dx`. `:.2I=pi*int_(0)^(pi)(sinx)/((1+cos^(2)x))dx`. Now, put `cosx=t`. |
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| 131. |
`int_(-pi)^(pi)x^(4)sinxdx=?`A. `2pi`B. `pi`C. `0`D. none of these |
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Answer» `f(x)=x^(4)sinx`. `:.f(-x)=(-x)^(4)sin(-x)=-(x^(4)sinx)=-f(x)` `:.f(x)` is odd and hence `I=0`. |
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| 132. |
The value of `int _(0)^(2) (3^sqrt(x))/(sqrt(x)) ` dx isA. `(2)/(log3)(3^(sqrt2)-1)`B. 0C. `2(sqrt2)/(log3)`D. `(3^(sqrt2))/(sqrt2)` |
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Answer» Correct Answer - A Put ` sqrt(x) = t rArr 1/(sqrt(x)) dx = 2 dt ` Also , as x = 0 to 2 so , `t = 0 "to" sqrt(2)` ` :. int _(0)^(2) (3^(sqrt(x)))/(sqrt(x)) dx = 2 int _(0)^(sqrt(2)) 3^(t) dt = 2 [ (3^(t))/(log 3)]_(0)^(sqrt(2)) = 2/(log 3 ) (3 ^(sqrt(2)) -1)` |
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| 133. |
` int _(0)^(pi//4) sec^(7) theta sin^(3) theta d theta ` is equal toA. `(1)/(12)`B. `(3)/(12)`C. `(5)/(12)`D. None of these |
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Answer» Correct Answer - C Let ` l = int _(0)^(pi//4) sec^(7) theta . d theta = int _(0)^(pi//4) (sin^(3) theta )/(cos^(3) theta ) theta d theta ` Put `tan theta = t rArr sec^(2) theta d theta = dt` ` :. l = int _(0)^(1) t^(3) (1+t^(2)) dt = [ (t^(4))/4+ (t^(6))/6 ] _(0)^(1) = 5/12 ` |
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| 134. |
`int_(-1)^(-2)x^(3)(1-x^(2))dx=?`A. `-(40)/(3)`B. `(40)/(3)`C. `(5)/(6)`D. `0` |
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Answer» `f(x)=x^(3)(1-x^(2))=(x^(3)-x^(5))`. `f(-x)=(-x)^(3)-(-x)^(5)=-x^(3)-(-x^(5))=-x^(3)+x^(5)=-x^(3)(1-x^(2))=-f(x)`. `:.f(x)` is odd and hence `I=0`. |
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| 135. |
Prove that `int_(0)^(4){|x|+|x-2|+|x-4|dx}=20`. |
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Answer» `I=int_(0)^(2){|x|+|x-2|+|x-4|}dx+int_(2)^(4){|x|+|x-2|+|x-4|}dx` `=int_(0)^(2){x-(x-2)-(x-4)}dx+int_(2)^(4){x+(x-2)-(x-4)}dx` `=int_(0)^(2)(x-x+2-x+4)dx+int_(2)^(4)(x+x-2-x+4)dx` `=int_(0)^(2)(-x+2)dx+int_(2)^(4)(x+2)dx`. |
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| 136. |
`int_(0)^(sqrt(2))sqrt(2-x^(2))dx=?` |
| Answer» Put `x=sqrt(2)sint`. | |
| 137. |
`int_(0)^(pi)(x tanx)/((secxcosecx))dx=(pi^(2))/(4)` |
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Answer» `I=int_(0)^(pi)xsin^(2)xdx` and `I=int_(0)^(pi)(pi-x)sin^(2)(pi-x)dx=int_(0)^(pi)(pi-x)sin^(2)xdx`. `:.2I=pi*int_(0)^(pi)sin^(2)xdx=(pi)/(2)*int_(0)^(pi)/(2)(1-cos2x)dx`. |
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| 138. |
` int _(0)^(a) (xdx)/(sqrt(a^(2)+x^(2)) )` is equal toA. `a(sqrt2-1)`B. `a(1-sqrt2)`C. `a(1+sqrt2)`D. `2asqrt3` |
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Answer» Correct Answer - A Put ` t = a^(2) +x^(2) rArr 2xdx = dt` ` :. int _(0)^(a) (xdx)/(sqrt(a^(2)+x^(2)) )= 1/2 int _(a^(2))^(2a^(2)) 1/(sqrt(t)) dt ` ` = [ ( 2a^(2) )^(1/2) - a^(2/2)]= a (sqrt(2)-1)` |
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| 139. |
`int_(-pi)^(pi)x^(3)cos^(3)xdx=?`A. `pi`B. `(pi)/(4)`C. `2pi`D. `0` |
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Answer» `f(x)=x^(3)cos^(3)x`. `:.f(-x)=(-x)^(3)[cos(-x)]^(3)=-x^(3)(cosx)^(3)=-(x^(3)cos^(3)x)=-f(x)` `:.f(x)` is odd and hence `I=0`. |
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| 140. |
`int_(-pi)^(pi)sin^(5)xdx=?`A. `(3pi)/(4)`B. `2pi`C. `(5pi)/(16)`D. `0` |
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Answer» `f(x)=sin^(5)x`. `f(-x)=[sin(-x)]^(5)=(-sinx)^(5)=-(sin^(5)x)=-f(x)`. `:.f(x)` is odd and hence `I=0`. |
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| 141. |
`int_0^2sqrt(6x+4)dx`A. `(64)/(9)`B. `7`C. `(56)/(9)`D. `(60)/(9)` |
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Answer» Correct Answer - C `I=int_(0)^(2)(6x+4)^(1//2)dx=[(2)/(3)*(6x+4)^(3//2)/(6)]_(0)^(2)=(56)/(9)`. |
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| 142. |
`int_(0)^(a)(x)/(sqrt(a^(2)+x^(2)))dx` |
| Answer» Correct Answer - Put `x=atantheta` | |
| 143. |
`int_(0)^(1)|3x-1|dx=` |
| Answer» `int_(0)^(1)|3x-1|=int_(0)^(1//3)(1-3x)dx+int_(1//3)^(1)(3x-1)dx=I_(1)+I_(2)` | |
| 144. |
`int_(0)^(pi/4) (sec x)/(1+2 sin^(2)x)` is equal toA. `(1)/(3)[log(sqrt2+1)+(pi)/(2sqrt2)]`B. `(1)/(3)[log(sqrt2+1)-(pi)/(2sqrt2)]`C. `3[log(sqrt2+1)-(pi)/(2sqrt2)]`D. `3[log(sqrt2+1)+(pi)/(2sqrt2)]` |
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Answer» Correct Answer - A Let `I=int_(0)^(pi//4)(cosx)/(cos^(2)x(1+2sin^(2)x))dx` `=int_(0)^(pi//4)(cosxdx)/((1-sin^(2)x)(1+2sin^(2)x))` Out `t=sin x rArr dt = cos x dx` `therefore I=int_(0)^(1//sqrt2)(1)/((1-t^(2))(1+2t^(2)))dt` `=(1)/(3)int_(0)^(1//sqrt2)((1)/(1-t^(2))+(2)/(1+2t^(2)))dt` `=(1)/(3)[(1)/(2.1)log((1+t)/(1-t))+(2)/(sqrt2)tan^(-1)tsqrt2]_(0)^((1)/(sqrt2))` `=(1)/(3)[(1)/(2)log((sqrt2+1)/(sqrt2-1))+sqrt2tan^(-1)1]` `=(1)/(3)[(1)/(2)log(sqrt2+1)^(2)+sqrt2.(pi)/(4)]=(1)/(3)[log(sqrt2+1)+(pi)/(2sqrt2)]` |
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| 145. |
`int_0^1sin^(- 1)((2x)/(1+x^2))dx` |
| Answer» Put `x=tant` and integrate by parts. | |
| 146. |
`int_0^pi xsin^2x dx=(pi^2)/4` |
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Answer» `I=int_(0)^(pi)xcos^(2)xdx` and `I=int_(0)^(pi)(pi-x)cos^(2)(pi-x)dx=int_(0)^(pi)(pi-x)cos^(2)xdx`. `:.2I=piint_(0)^(pi)cos^(2)xdx=(pi)/(2)*int_(0)^(pi)/(2)(1+cos2x)dximpliesI=(pi^(2))/(4)`. |
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| 147. |
`int_(1)^(4)xsqrt(x)dx=?`A. `12.8`B. `12.4`C. `7`D. none of these |
| Answer» `I=int_(1)^(4)x^(3//2)dx=[(2)/(3)x^(5//2)]_(1)^(4)=(62)/(5)=12.4`. | |
| 148. |
` int _(0)^(pi//6) (2+3x^(2))cos 3 xdx ` is equal toA. `(1)/(36)(pi+16)`B. `(1)/(36)(pi-16)`C. `(1)/(36)(pi^(2)-16)`D. `(1)/(36)(pi^(2)+16)` |
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Answer» Correct Answer - D ` int _(0)^(pi//6) (2+3x^(2) ) cos 3x` ` = [ (sin 3x)/3 (2+3x^(2)) ]_(0)^(pi/6) - int _(0)^(pi//6) (sin 3x)/3 . 6x dx ` ` = 1/36 (pi^(2)+16)` |
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| 149. |
`int_(0)^(2)xsqrt(2-x)dx` |
| Answer» `I=int_(0)^(2)(2-x)[2-(2-x)]^(1//2)dx=int_(0)^(2)(2-x)sqrt(x)dx=int_(0)^(2)(2sqrt(x)-x^(3//2))dx`. | |
| 150. |
`int_(0)^(2)xsqrt(2+x)dx` |
| Answer» Correct Answer - Put `(x+2)=t^(2)`. | |