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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 x sqrt(x+2)dx` is equal toA. `(2)/(5)(x+2)^(5//2)-(4)/(3)(x+2)^(3//2)+C`B. `(1)/(5)(x+2)^(5//2)-(4)/(3)(x+2)^(3//2)+C`C. `(2)/(5)(x+2)^(5//2)-(2)/(3)(x+2)^(3//2)+C`D. None of these |
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Answer» Correct Answer - A `intxsqrt(x+2)dx=int(x+2-2)sqrt(x+2)dx` `=int(x+2)sqrt(x+2)dx-2intsqrt(x+2)dx` `=int(x+2)^(3//2)dx-2int(x+2)^(1//2)dx` `=((x+2)^((3//2)+1))/((3//2)+1)-2((x+2)^((1//2)+1))/((1//2)+1)+C` `" "[because intx^(n)dx=(x^(n+1))/(n+1)]` `=(2)/(5)(x+2)^(5//2)-(2xx2)/(3)(x+2)^(3//2)+C` `=(2)/(5)(x+2)^(5//2)-(4)/(3)(x+2)^(3//2)+C` |
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| 102. |
`int(sinx)/(sinx-a)dx` is equal toA. `x+sin a cos a log|sin(x-a)|+C`B. `(cosa)x+sin alog|sin(x-a)|+C`C. `(sina)x+cosalog|sin(x-a)|+C`D. None of the above |
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Answer» Correct Answer - B `int(sinx)/(sin(x-1))dx=int(sin{(x-a)+a})/(sin(x-1))dx` `=int(sin(x-a)cosa+cos(x-a)sina)/(sin(x-a))dx` `=int(sin(x-a)cosa)/(sin(x-a))dx+int(cos(x-a)sina)/(sin(x-a))dx` `=cosa int 1dx+sin a int cot(x-a)dx` `=(cosa)x+sina.log|sin(x-a)|+C` |
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| 103. |
`int(x)/(sqrt(1+x^(2)))dx` |
| Answer» Correct Answer - `(1)/(5) (1-x^(3))^(5//3) -(1)/(5) (1-x^(3))^(2//3) +c` | |
| 104. |
`int " log sin x . sec"^(2) " x dx "` |
| Answer» Correct Answer - `tan x log (sin x)-x+c` | |
| 105. |
`int(1)/(sqrt(1+x-x^(2)))dx` |
| Answer» Correct Answer - `sin^(-1) ((2x-1)/(sqrt(5)))+c` | |
| 106. |
`int(1-cosx)cosec^(2)x dx` is equal toA. `tan x+C`B. `tan.(x)/(2)+C`C. `(1)/(2)tan.(x)/(2)+C`D. None of these |
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Answer» Correct Answer - B `int(1-cosx)"cosec"^(2)xdx` `=int"cosec"^(2)xdx-int"cosec"^(2)xcosxdx` `=-cotx+"cosec x"+C=(1-cosx)/(sinx)+C` `=(2sin^(2).(x)/(2))/(2sin.(x)/(2)cos.(x)/(2))+C` `=tan.(x)/(2)+C` |
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| 107. |
`intx^(2) . tan^(2) x^(3). sec^(2) x^(3) dx` |
| Answer» Correct Answer - `(1)/(12) tan^(4) x^(3) +c` | |
| 108. |
`int(1)/(sqrt2+x-3x)dx` |
| Answer» Correct Answer - `(1)/(sqrt(3)) sin^(-1) ((6x-1)/(5))+c` | |
| 109. |
`int (x) /(sqrt(1+x^(4)) ) dx` |
| Answer» Correct Answer - `(1)/(2) sin^(-1) x^(2) +c` | |
| 110. |
`int"cosec"(x-a)"cosec"xdx` is equal toA. `(-1)/(sina)log|sinx"cosec "(x-a)|+C`B. `(-1)/(sina)log[sin(x-a)sinx]+C`C. `(1)/(sina)log[sin(x-a)"cosec x"]+C`D. `(1)/(sina)log[sin(x-a)sinx]+C` |
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Answer» Correct Answer - A Let `l=int"cosec"(x-a)" cosec" x dx` `=int(sin a)/(sin a sin(x-a)sinx)dx` `=-(1)/(sina)int(sin[(x-a)-x])/(sin(x-a)sin)dx` `=-(1)/(sina) int[(sin(x-a)cosx - cos(x-a)sinx)/(sin(x-a)sinx)]dx` `=-(1)/(sina)int[cot x - cot(x-a]dx` `=-(1)/(sina)[log|sinx |-log|sin(x-a)|]+C` `=-(1)/(sina)[log||sinx" cosec"(x-a)|]+C` |
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| 111. |
Evaluate `int(1-cosx)"cosec"^(2)xdx`A. `tanx+C`B. `tan.(x)/(2)+C`C. `(1)/(2)tan.(x)/(2)+C`D. None of these |
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Answer» Correct Answer - B `int(1-cosx)"cosec"^(2)xdx` `=int"cosec"^(2)xdx-int"cosec"^(2)xcosxdx` `=-cotx+"cosec x"+C=(1-cosx)/(sinx)+C` `=(2sin^(2).(x)/(2))/(2sin.(x)/(2)cos.(x)/(2))+C` `=tan.(x)/(2)+C` |
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| 112. |
`int(x^(3))/((4-x^(4))^(2))dx` |
| Answer» Correct Answer - `(1)/(4(4-x^(4)))+c` | |
| 113. |
`intx cos^(3) x^(2) . sin x^(2) dx` |
| Answer» Correct Answer - `-(1)/(8)cos^(4) x^(2) +c` | |
| 114. |
`int(dx)/(sinx-cosx+sqrt2)` is equal toA. `-(1)/(sqrt2)tan((x)/(2)+(pi)/(8))+C`B. `(1)/(sqrt2)tan((x)/(2)+(pi)/(8))+C`C. `(1)/(sqrt2)cot((x)/(2)+(pi)/(8))+C`D. `-(1)/(sqrt2)cot((x)/(2)+(pi)/(8))+C` |
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Answer» Correct Answer - D `int(dx)/(sinx-cosx+sqrt2)` `=int(dx)/(sqrt2((1)/(sqrt2)sinx-(1)/(sqrt2)cosx)+sqrt2)` `=(1)/(sqrt2)int(dx)/(1-cos (x+(pi)/(4)))=(1)/(sqrt2)int(dx)/(2sin^(2)((x)/(2)+(pi)/(8)))` `=(1)/(2sqrt2)int(dx)/(sin^(2)((x)/(2)+(pi)/(8)))=(1)/(2sqrt2)int"cosec"^(2)((x)/(2)+(pi)/(8))dx` `=(1)/(2sqrt2)[(-cot ((x)/(2)+(pi)/(8)))/((1)/(2))]+C=-(1)/(sqrt2)cot((x)/(2)+(pi)/(8))+C` |
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| 115. |
`int e^(3x) " cos 2x dx "` |
| Answer» Correct Answer - `(e^(3x))/(13) ( 3 sin 2x +3 cos 2x)+c` | |
| 116. |
`int (x^(2) +1)/(sqrt(x^(2)+3))dx` |
| Answer» Correct Answer - `(1)/(2)[xsqrt(x^(2)+3)-log |x+sqrt(x^(2)+3)]+c` | |
| 117. |
Evaluate:`intx^2sqrt(a^6-x^6)dx` |
| Answer» Correct Answer - `(1)/(6) [x^(3) sqrt(a^(6)-x^(6))+a^(6) sin^(-1) ((x^(3))/(a^(3))) ]+c` | |
| 118. |
`intx^2sqrt(x^6-1)dx` |
| Answer» Correct Answer - `(1)/(6)[x^(3)sqrt(x^(6)-1)-log |x^(3)+sqrt(x^(6)-1)|]+c` | |
| 119. |
`intsqrt(1-4x-x^(2))dx`A. `(x)/(2)sqrt(1-4x-x^(2))+(5)/(2)sin^(-1)((x+2)/(sqrt5))+C`B. `(x)/(2)sin^(-1)((x+2)/(sqrt5))+(5)/(2)sqrt(1-4x-x^(2))+C`C. `(x+2)/(2)sin^(-1)((x+2)/(sqrt5))+(5)/(2)sqrt(1-4x-x^(2))+C`D. `(x+2)/(2)sqrt(1-4x-x^(2))+(5)/(2)sin^(-1)((x+2)/(sqrt5))+C` |
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Answer» Correct Answer - D Let `l=intsqrt(1-4x-x^(2))dx` `=intsqrt(-(x^(2)+4x-1-2^(2)+2^(2)))dx` `=int sqrt(-[(x+2)^(2)-(sqrt5)^(2)])dx` `=intsqrt((sqrt5)^(2)-(x+2)^(2))dx` `[because int sqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(A^(2)-x^(2))+(a^(2))/(2)sin^(-1).(x)/(a)+C]` `rArr " "l=(x+2)/(2)sqrt(1-4x-x^(2))+(5)/(2)sin^(-1).((x+2))/(sqrt5)+C` |
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| 120. |
`int(dx)/(e^x+e^(-x))`is equal toA. `tan^(-1) (e^(x)) +C`B. `log(e^(x)-e^(-x))+C`C. `log (e^(x)+e^(-x))+C`D. `tan^(-1) (e^(2x)) +C` |
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Answer» Correct Answer - A `I= int(1)/(e^(x)+e^(-x))dx= int(1)/(e^(x)+(1)/(e^(x)))dx` `=(e^(x))/((e^(x))^(2)+1)dx" " underset(rArr e^(x) dx=dt)(" Let " e^(x)=t)` `:. I= int(1dt)/(t^(2)+1)=tan^(-1) (t)+C =tan^(-1)e^(x)+C` |
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| 121. |
`int(cos 2x)/((sin x + cos x)^(2)) dx " is equal to : "`A. `log|sin x+cos x|+C`B. `log|sin x-cos x|+C`C. `(1)/((sin x+ cos x)^(2))`D. |
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Answer» Correct Answer - b `" Let I "= int(cos2x)/((sin x + cos x)^(2))dx` `=int(cos^(2) x-sin^(2)x)/((sin x+cos x)^(2))dx` ` =int((cos x-sin x)(cos x+ sin x))/((sin x+cos x)(sin x+cos x))dx` `=int( cos x- sinx)/(sin x+cos x)dx` `" Let " sin x cos x=t rArr (cos x-sin x) dx=dt` `:. I= int (1)/(t) dt= log|t|+C` `=log |sin x+cos x| +C` |
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| 122. |
`int(x+sinx)/(1+cosx) dx` is equal toA. `log|1+cosx|+C`B. `log|x+sinx|+C`C. `x-tanx//2 +C`D. `x tan((x)/(2))+C` |
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Answer» Correct Answer - D Let `l=int(x+sinx)/(1+cosx)dx` `int(x)/(2cos^(2)x//2)dx+int(2sin x//2cos x//2)/(2cos^(2)x//2)dx` `=(1)/(2)intx sec^(2).(x)/(2)dx+int tan.(x)/(2)dx` On integrating by parts in first integral, we get `=(1)/(2)[x int sec^(2).(x)/(2)dx-int[(d)/(dx)(x)intsec^(2).(x)/(2)dx]dx]+inttan.(x)/(2)dx` `=(1)/(2)[x tan.(x)/(2).2-2 int tan.(x)/(2)dx]+int tan.(x)/(2)dx` `=x tan .(x)/(2)-int tan.(x)/(2)dx+int tan.(x)/(2)dx` `=x tan.(x)/(2)+C` |
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| 123. |
Choose the correct answer`intsqrt(x^2-8x+7)dx`(A) `1/2(x-4)sqrt(x^2-8x+7)+9log|x-4+sqrt(x^2-8x+7)|+C`(B) `1/2(x+4)sqrt(x^2-8x+7)+9log|x+4+sqrt(x^2-8x+7)|+C`(C) `1/2(x-4)sqrt(x^2-8x+7)-3sqrt(2)log|x-4+sqrt(x^2-8x+7)|+C`(D) `1/2(x-4)sqrt(x^2-8x+7)-A. `(1)/(2)(x+4)sqrt(x^(2)-8x+7)` `+9log |x+4+sqrt(x^(2)-8x+7|+C`B. `(1)/(2)(x-4)sqrt(x^(2) -8x+7)` `-3sqrt(2)log|x-4+sqrt(x^(2)-8x+7|+C`C. `(1)/(2) (x-4)sqrt(x^(2)-8x+7)` `-(9)/(2)log|x-4+sqrt(x^(2)-8x+7|+C`D. |
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Answer» Correct Answer - D `" Let " I= int sqrt(x^(2)-8x+7)dx` `rArr I=int sqrt(x^(2)-8x+7+(4)^(2)-(4)^(2))dx` `= int sqrt((x-4)^(2)+7-16)dx` `=int sqrt((x-4)^(2)-(3)^(2))dx` `rArr I=(x-4)/(2)sqrt(x^(2)-8x+7)` `-(9)/(2)log |x+4+sqrt(x^(2)-8x+7)|+c` |
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| 124. |
`int(((x^(2)+2)a^((x+tan^(-1)x)))/(x^(2)+1))dx` is equal toA. `log a.a^(x+tan^(-1)x)+c`B. `((x+tan^(-1)x))/(loga)+c`C. `(a^(x+tan^(-1)x))/(loga)+c`D. `loga(x+tan^(-1)x)+c` |
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Answer» Correct Answer - C Let `l=int((x^(2)+2)a^((x+tan^(-1)x)))/(x^(2)+1)dx` `"Put "x+tan^(-1)x=t` `rArr" "(1+(1)/(1+x^(2)))dx=dt` `rArr" "(2+x^(2))/(1+x^(2))dx=dt` `therefore" "l=inta^(t)dt=(a^(t))/(loga)+c=(a^(x+tan^(-1)x))/(loga)+c` |
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| 125. |
`int(1)/(sqrt(9x-4x^(2)))dx` is equal toA. `(1)/(9)sin^(-1)((9x-8)/(8))+C`B. `(1)/(2)sin^(-1)((8x-9)/(9))+C`C. `(1)/(3)sin^(-1)((9x-8)/(8))+C`D. `(1)/(2)sin^(-1)((9x-8)/(9))+C` |
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Answer» Correct Answer - B `int(1)/(sqrt(9x-4x^(2)))dx=(1)/(sqrt4)int(1)/(sqrt((9)/(4)x-x^(2)))dx` `=(1)/(2)int(1)/(sqrt(-[x^(2)-(9)/(4)x+((9)/(8))^(2)]+((9)/(8))^(2)))dx` `=(1)/(2)int(1)/(sqrt(((9)/(8))^(2)-(x-(9)/(8))^(2)))dx` `=(1)/(2)sin^(-1)((x-(9)/(8))/((9)/(8)))+C` `=(1)/(2)sin^(-1)((8x-9)/(9))+C` |
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| 126. |
Choose the correct answer`int1sqrt(3)(dx)/(1+x^2)`equals(A) `pi/3`(B)`(2pi)/3`(C)`pi/6`(D)`pi/(12)`A. `(2pi)/(3)`B. `(pi)/(6)`C. `(pi)/(12)`D. |
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Answer» Correct Answer - d `int_(1)^(sqrt(3))(1)/(1+x^(2))dx` `=[tan^(-1)x]_(1)^(sqrt(3))` `=tan^(-1)sqrt(3)-tan^(-1)1` `=(pi)/(3)-(pi)/(4)rArr (4pi-3pi)/(12)=(pi)/(12)` |
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| 127. |
`int sqrt(2-3x^(2))dx` |
| Answer» Correct Answer - `(x)/(2) sqrt(2-3x^(2))+(1)/(sqrt(3)) sin^(-1) ((xsqrt(3))/(sqrt(2))) +c` | |
| 128. |
`(i) int(tan^(-1))/((1+x^(2)))dx" "(ii) int(1)/(sqrt(1-x^(2)) sin^(-1)x)dx` |
| Answer» Correct Answer - `(i) (1)/(2) (tan^(-1)x)^(2) +c (ii) log (sin^(-1)x) +c` | |
| 129. |
`(i) int(e^(x))/(1+e^(x))dx" "(ii) int (e^(x)) /((1+e^(x))^(4))dx` |
| Answer» Correct Answer - `(i) log |1+e^(x)|+c (ii) -(1)/(3(1+e^(x))^(3))+c` | |
| 130. |
If `int(f(x))/(log(sinx))dx=log[log sinx]+c`, then f(x) is equal toA. `cot x`B. `tanx`C. `sec x`D. `"cosec x"` |
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Answer» Correct Answer - A Given, `int(f(x))/(log(sinx))dx=log[logsinx]+c` On differentiating both sides, we get `(f(x))/(log(sinx))=(1)/(log sinx)(d)/(dx)(log sinx)+0` `rArr" "(f(x))/(log(sinx))=(1)/(log sinx)xx(1)/(sinx)xx(1)/(sinx)xxcosx` `rArr" "f(x)=cotx` |
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| 131. |
`int (2x+3)/((x+2)(x-2))dx` |
| Answer» Correct Answer - `(1)/(4) log |x+2|+(7)/(4) log |x-2|+c` | |
| 132. |
`int (x)/((x-2)(x+1))dx` |
| Answer» Correct Answer - `(1)/(3) log |(x-2)^(2)(x+1)|+c` | |
| 133. |
Choose the correct answer`int0 2/3(dx)/(4+9x^2)`(A) `pi/6`(B)`pi/(12)`(C)`pi/(24)`(D)`pi/4`A. `(pi)/(12)`B. `(pi)/(24)`C. `(pi)/(4)`D. |
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Answer» Correct Answer - c `int_(0)^(2//3)(1)/(4+9x^(2))dx` `=(1)/(9)int_(0)^(2//3)(1)/((4)/(9)+x^(2))dx` `=(1)/(9)int_(0)^(2//3)(1)/(((2)/(3))^(2)+x^(2))dx` `=(1)/(9).(1)/(2//3)[tan^(-1)((x)/(2//3))]_(0)^(2//3)` `=(1)/(6)[tan^(-1)((3x)/(2))]_(0)^(2//3)` `=(1)/(6)[tan^(-1)((3)/(2).(2)/(3))-tan^(-1)0]` `=(1)/(6) (tan^(-1)1-0)=(1)/(6).(pi)/(4)=(pi)/(24)` |
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| 134. |
Choose the correct answerThe value of the integral `int1/3 1((x-x^3)^(1/3))/(x^4)dx`is(A) 6 (B)0 (C) 3 (D) 4 |
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Answer» Correct Answer - a `"Let I "=int_(1//3)^(1)((x-x^(3))^(1/3))/(x^(4))dx` `=int_(1//3)^(1)((x^(3))^(1//3)((x)/(x^(3))-(x^(3))/(x^(3)))^(1//3))/(x^(4))dx` `=int_(1//3)^(1)(((1)/(x^(2))-1))/(x^(3))dx` `" Let " (1)/(x^(2))=t rArr (-2)/(x^(3)) dx=dt rArr (dx)/(x^(3)) =(dt)/((-2))` `x=1 rArr t=1 ` ` "and " x=(1)/(3) rArr t=3^(2)=9` `:. I= int_(9)^(1) (t-1)^(1//3) (dt)/((-2)) =(1)/(2) [((t-1)^(1/3+1))/((1)/(3)+1)]_(9)^(1)` `= -(1)/(2) xx(3)/(4) [(t-1)^(4/3)]_(9)^(1) =-(3)/(8)[(1-1)^(4/3)-(9-1)^(4/3)]` `=-(3)/(8)[0-(2^(3))^(4/3)]=-(3)/(8)xx(-16)=6` |
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| 135. |
Evaluate: (i) `int1/(a^2-b^2 x^2) dx`(ii) `int1/(a^2 x^2-b^2) dx` |
| Answer» Correct Answer - `(1)/(2 ab) log |(ax-b)/(ax+b)| +c` | |
| 136. |
`int(cosx-1)/(sinx+1).e^(x)dx` is equal toA. `(e^(x)cosx)/(1+sinx)+C`B. `C-(e^(x)sinx)/(1+sinx)`C. `C-(e^(x))/(1+sinx)`D. `C-(e^(x)cosx)/(1+sinx)` |
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Answer» Correct Answer - A Let `l=int(cosx)/(1+sinx).e^(x)dx-int(1)/(sinx+1).e^(x)dx` integration by parts in first integral, we get `((cosx)e^(x))/(1+sinx)-int[(-(1+sinx)sinx-cos^(2)x)/((1+sinx)^(2))].e^(x)dx-int(e^(x))/(sinx+1)dx` `=(e^(x)cosx)/(1+sinx)+int(1)/(1+sinx).e^(x)dx-int(e^(x)dx)/(1+sinx)` `=(e^(x)cosx)/(1+sinx)+C` |
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| 137. |
`int e^(2x) " (tan x+1)"^(2) dx` |
| Answer» Correct Answer - `e^(2x) . Tan x+c` | |
| 138. |
`int(sinx. cos x)/(a^(2) cos^(2) x+b^(2) sin^(2) x)dx` |
| Answer» Correct Answer - `(1)/(2(b^(2)-a^(2))) log _(e) (a^(2) cos^(2) x+b^(2) sin^(2)x) +c` | |
| 139. |
The value of `int(dx)/(x(x^(n)+1))` is equal toA. `(1)/(n)log((x^(n))/(x^(n)+1))+C`B. `log((x^(n)+1)/(x^(n)))+C`C. `(1)/(n)log((x^(n)+1)/(x^(n)))+C`D. `log((x^(n))/(x^(n)+1))+C` |
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Answer» Correct Answer - A Let `l=int(dx)/(x(x^(n)+1))(" let "t=x^(n)+1, dt=nx^(n-1)dx)` `=int(dt)/(nx^(n).t)" "[(dt)/(nx^(n))=(dx)/(x)]` `=(1)/(n)int(dt)/(t(t-1))=(1)/(n)int{(1)/(t-1)-(1)/(t)}dt` `=(1)/(n){log(t-1)-logt}+C` `=(1)/(2)log(t-1)/(t)+C` `=(1)/(n)log(t-1)/(t)+C=(1)/(n)log((x^(n))/(x^(n)+1))+C` |
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| 140. |
`int(cotx)/(log(sinx)dx` |
| Answer» Correct Answer - `log |log sin x|+c` | |
| 141. |
The value of `int cos (logx)dx` isA. `(1)/(2)[sin (logx)+cos (logx)]+C`B. `(x)/(2)[sin(logx)+cos(logx)]+C`C. `(x)/(2)[sin(logx)-cos (logx)]+C`D. `(1)/(2)[sin(logx)-cos (logx)]+C` |
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Answer» Correct Answer - B Let `" "l=int cos (logx).1dx" …(i)"` Use integral by parts, `l=cos(logx).x -int[-sin (logx)].(1)/(x).xdx` `=x.cos(logx)+int sin (logx).1dx` Again on using integration by part, `=x-cos(logx)+[sin(logx).x-int cos(logx).(1)/(x).xdx]+C` `=x.cos(logx)+[x sin(logx)-int cos(logx)dx]+C` `=x {sin(logx)+cos(logx)}-l+C" [from Eq. (i)]"` `rArr" "l=(x)/(2){sin (logx)+cos(logx)}+C` |
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| 142. |
The value of `inte^(x)(x^(5)+5x^(4)+1)dx` isA. `e^(x).x^(5)+C`B. `e^(x).x^(5)+e^(x)+C`C. `e^(x+1).e^(5)+C`D. `5x^(4).e^(x)+C` |
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Answer» Correct Answer - B Let `l=int e^(x)(x^(5)+5x^(4)+1)dx` `=int e^(x)x^(5)dx+5inte^(x)x^(4)dx+inte^(x)dx` integration by parts, we get `x^(5)e^(x)-int 5x^(4)e^(x)dx+5 int e^(x)x^(4)dx+e^(x)=x^(5)e^(x)+e^(x)+C` |
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| 143. |
`int (x-2)/(x^(3)).e^(x) dx` |
| Answer» Correct Answer - `(e^(x))/(x^(2))+c` | |
| 144. |
`int(sinx)/(a+b cos x)dx` |
| Answer» Correct Answer - `-(1)/(b) log|a+bcos x| +c` | |
| 145. |
`int(logx/(1+logx)^2)dx` |
| Answer» Correct Answer - `(x)/(1+log x)+c` | |
| 146. |
`int (1)/(x-x^(3))dx` |
| Answer» Correct Answer - `log|x|-(1)/(2) log |1-x|-(1)/(2) log |1+x|+c` | |
| 147. |
If `f(x)=(sin^(-1)x)/(sqrt(1-x^(2))) and g(x)=e^(sin^(-1)x),` then `int f(x)g(x)dx` is equal toA. `e^(sin^(-1)x)(sin^(-1)x-1)+C`B. `e^(sin^(-1)x)+C`C. `e^((sin^(-1)x)^(2))+C`D. `e^(2sin^(-1)x)+C` |
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Answer» Correct Answer - A `intf(x)g(x)=int(sin^(-1)x)/(sqrt(1-x^(2)))e^(sin^(-1))dx` Put`" "sin^(-1)x=t` `rArr" "(1)/(sqrt(1-x^(2)))dx=dt` `therefore" "intf(x)g(x)dx=int te^(t)dt=te^(t)-e^(t)+C` `=e^(sin^(-1)x)(sin^(-1)x-1)+C` |
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| 148. |
`inte^(x)(1-cotx+cot^(2)x)dx=`A. `e^(x)cotx+C`B. `e^(x)"cosec x"+C`C. `-e^(x)cotx+C`D. `-e^(x)"cosec x"+C` |
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Answer» Correct Answer - C `inte^(x)(1-cot x+cot^(2)x)dx` `=int e^(x)(-cot x+"cosec"^(2)x)dx` `=-int e^(x) cot x dx +int e^(x)" cosec"^(2)xdx` On integration by parts, we get `-e^(x) cot x - int e^(x)" cosec"^(2) x dx +int e^(x)"cosec"^(2) dx` `=-e^(x)cotx+C` |
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| 149. |
`int(1)/(2x^(2)+x+1)dx` |
| Answer» Correct Answer - `(2)/(sqrt(7)) tan^(-1) ((4x+1)/sqrt(7))+c` | |
| 150. |
`inttan^(-1)xdx=….+C`A. `(1)/(1+x^(2))`B. `x tan^(-1)x+(1)/(2)log|1+x^(2)|`C. `x tan^(-1)x+(1)/(2).(tan^(-1)x)/(1+x^(2))`D. `x tan^(-1)x-(1)/(2)log|1+x^(2)|` |
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Answer» Correct Answer - D Let `l=int tan^(-1)underset("I")(x).underset("II")(1)dx` On integration by parts, we get `tan^(-1)x.x-int(1)/(1+x^(2)).x dx = x tan^(-1)-l_(1)" …(i)"` where, `" "l_(1)=int(x)/(1+x^(2))dx` Put `1+x^(2)=t rArr x dx=(1)/(2)dt` `therefore" "l_(1)=(1)/(2)int (1)/(t)dt=(1)/(2)log t=(1)/(2)log(1+x^(2))` On putting the value of `l_(1)` in Eq. (i), we get `l=x tan^(-1)x-(1)/(2)log(1+x^(2))+C` |
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