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AAKASH SERIES-DEFINITE INTEGRALS-EXERCISE - II
- For a gt 1, b gt 1, the value of int(0)^(oo) (a^(-x)-b^(-x))dx=
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- If int(ln 2)^(x) (e^(x)-1)^(-1)dx= ln (3/2) then x=
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- int(0)^(log5)(e^(x)sqrt(e^(x)-1))/(e^(x)+3)dx=
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- int(2)^(3) (1)/(x^(2)-x) dx=
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- int(0)^(3) (3x + 1)/(x^(2)+9)dx=
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- int(0)^(1) (dx)/(x^(2)+2x sin alpha +1)=
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- int(0)^(pi//4) (sin^(2) x cos^(2) x)/((sin^(3) x + cos^(3)x)^(2))dx=
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- int(1//2)^(1) (1)/(sqrt(x-x^(2)))dx=
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- int(0)^(oo) (dx)/((x+sqrt(x^(2)+1))^(3))=
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- If int(0)^(b)(dx)/(1+x^(2))=int(b)^(oo)(dx)/(1+x(2)), then b=
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- int(0)^(pi//2) (8+7 cos x )/((7+8 cos x)^(2))dx is equal to
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- int(1//3)^(1) ((x-x^(3))^(1//3))/(x^(4))dx=
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- int(1)^(e) ((ln x )^(3))/(x) dx=
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- int(2)^(4) (sqrt(x^(2)-4))/(x^(4))dx=
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- int(0)^(1) sin (2 tan^(-1) sqrt((1+x)/(1-x)))dx =
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- int(0)^(pi//2) (Sinx+Cos x)/(3+ Sin 2x)dx=
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- int(0)^(pi//4) [sqrt(tan x)+sqrt(cot x)]dx
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- 2 int(0)^(1) (tan^(-1)x)/(x) dx=
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- If int(0)^(oo)(x^(2)dx)/((x^(2)+a^(2))(x^(2)+b^(2))(x^(2)+c^(2)))=(pi)...
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- int(0)^(pi) (1)/(3+2 sin x+ cos x)dx=
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