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AAKASH SERIES-INDEFINITE INTEGRALS -PRACTICE EXERCISE
- int (tan (sin^(-1)x))/(sqrt(1-x^(2)))dx=
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- int (Sin^(-1) x)/(sqrt(1-x^(2)))dx=
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- int (sqrt(x))/(1 + x) dx =
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- int ((1+x)e^(x))/(cos^(2)( xe^(x)))dx=
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- int (x^(3))/(1+x^(8))dx=
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- int (e^(x))/((2e^(x) -3)^(2)) dx =
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- int e^(sin^(2)x + cos x) (sin 2x-sin x)dx=
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- int e^(cot x + 2" log cosecx") dx =
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- int sqrt(tan x) sec^(2) x dx =
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- Evaluate the integerals. int (x^(2))/(sqrt(1-x))dx, x in (-oo, 1).
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- int (x^(5//2))/(sqrt(1 + x^(7))) dx =
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- int (sec^(2)x)/(sqrt(a+b tan x))dx=
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- int (dx)/(sqrt(sin^(3)x cos x))=g (x)+c rArr g(x)=
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- int (dx)/(cos x sqrt(cos 2x)) =
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- int (1)/(sin x sqrt(" sin x cos x")) dx =
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- int([x+sqrt(a^(2)+x^(2))]^(n))/sqrt(a^(2)+x^(2))dx(n ne0)=
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- int (secx)/(a cos x+ b sin x)dx=
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- Evaluate the integerals. int (1)/(x log x [log g (log x)])dx on (1...
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- int (1)/(1 - x^(2)) " ln " ((1 + x)/(1 -x)) dx =
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- int (1)/(x sqrt(x^(6) + 1)) dx =
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