01Hard64×since 2004Q4007If ∫tanx1+tanx+tan2xdx=x−KAtan−1\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}∫1+tanx+tan2xtanxdx=x−AKtan−1 (K tanx+1A)+C,(C \left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,(AKtanx+1)+C,(C is a constant of integration) then the ordered pair (K, A) is equal to :A(2, 1)B(−-−2, 3)C(2, 3)D(−-−2, 1)Check answerSkip
02Medium64×since 2004Q4008The integral ∫sin2xcos2x(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2dx\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx∫(sin5x+cos3xsin2x+sin3xcos2x+cos5x)2sin2xcos2xdx is equal toA−11+cot3x+C{{ - 1} \over {1 + {{\cot }^3}x}} + C1+cot3x−1+CB13(1+tan3x)+C{1 \over {3\left( {1 + {{\tan }^3}x} \right)}} + C3(1+tan3x)1+CC−13(1+tan3x)+C{{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + C3(1+tan3x)−1+CD11+cot3x+C{1 \over {1 + {{\cot }^3}x}} + C1+cot3x1+CCheck answerSkip
03Medium64×since 2004Q4009Let n ≥\ge≥ 2 be a natural number and 0<θ<π2.0 < \theta < {\pi \over 2}.0<θ<2π. Then ∫(sinnθ−sinθ)1/ncosθsinn+1θ dθ\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta∫sinn+1θ(sinnθ−sinθ)1/ncosθdθ is equal to - (where C is a constant of integration)Ann2−1(1+1sinn−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1+sinn−1θ1)nn+1+CBnn2−1(1−1sinn+1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1−sinn+1θ1)nn+1+CCnn2−1(1−1sinn−1θ)n+1n+C{n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2−1n(1−sinn−1θ1)nn+1+CDnn2+1(1−1sinn−1θ)n+1n+C{n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + Cn2+1n(1−sinn−1θ1)nn+1+CCheck answerSkip
04Hard64×since 2004Q4010Let a∈(0,π2)a \in \left( {0,{\pi \over 2}} \right)a∈(0,2π) be fixed. If the integral ∫tanx+tanαtanx−tanαdx\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx∫tanx−tanαtanx+tanαdx = A(x) cos 2α\alphaα + B(x) sin 2α\alphaα + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :Ax−αx - \alphax−α and loge∣cos(x−α)∣{\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|loge∣cos(x−α)∣Bx+αx + \alphax+α and loge∣sin(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge∣sin(x−α)∣Cx+αx + \alphax+α and loge∣sin(x+α)∣{\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|loge∣sin(x+α)∣Dx−αx - \alphax−α and loge∣sin(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge∣sin(x−α)∣Check answerSkip
05Easy64×since 2004Q4011If ∫x5e−x2dx=g(x)e−x2+c\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c∫x5e−x2dx=g(x)e−x2+c, where c is a constant of integration, then ggg(–1) is equal to :A1B- 1C−52- {5 \over 2}−25D−12- {1 \over 2}−21Check answerSkip
06Medium64×since 2004Q4012If ∫dx(x2−2x+10)2=A(tan−1(x−13)+f(x)x2−2x+10)+C\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C∫(x2−2x+10)2dx=A(tan−1(3x−1)+x2−2x+10f(x))+C where C is a constant of integration then :AA =154{1 \over {54}}541 and f(x) = 9(x–1)²BA =154{1 \over {54}}541 and f(x) = 3(x–1)CA =181{1 \over {81}}811 and f(x) = 3(x–1)DA =127{1 \over {27}}271 and f(x) = 9(x–1)²Check answerSkip
07Medium64×since 2004Q4013The integral ∫sec2/3 x cosec4/3x dx\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}}∫sec2/3xcosec4/3xdx is equal to (Hence C is a constant of integration)A-3/4 tan ^- 4 / 3 x + CB3tan^–1/3x + CC–3cot^–1/3x+ CD- 3tan^–1/3x + CCheck answerSkip
08Medium64×since 2004Q4014If ∫dxx3(1+x6)2/3=xf(x)(1+x6)13+C\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}∫x3(1+x6)2/3dx=xf(x)(1+x6)31+C where C is a constant of integration, then the function ƒ(x) is equal toA3x2{3 \over {{x^2}}}x23B−16x3- {1 \over {6{x^3}}}−6x31C−12x3- {1 \over {2{x^3}}}−2x31D−12x2- {1 \over {2{x^2}}}−2x21Check answerSkip
09Easy64×since 2004Q4016If ∫x+12x−1 dx\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx∫2x−1x+1dx = f(x) 2x−1\sqrt {2x - 1}2x−1 + C, where C is a constant of integration, then f(x) is equal to :A23{2 \over 3}32 (x −-− 4)B13{1 \over 3}31 (x + 4)C13{1 \over 3}31 (x + 1)D23{2 \over 3}32 (x + 2)Check answerSkip
10Medium64×since 2004Q4017If ∫1−x2x4\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}∫x41−x2 dx = A(x)(1−x2)m{\left( {\sqrt {1 - {x^2}} } \right)^m}(1−x2)m + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))^m equals :A127x6{1 \over {27{x^6}}}27x61B−127x9{{ - 1} \over {27{x^9}}}27x9−1C19x4{1 \over {9{x^4}}}9x41D13x3{1 \over {3{x^3}}}3x31Check answerSkip