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AAKASH SERIES-DEFINITE INTEGRALS-PRACTICE EXERCISE
- int(0)^(pi//4)[sqrt([1-sin2x)/(1+sin2x)]dx=
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- int(-1)^(1)log((2+x)/(2-x))dx=
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- int(-1)^(1)(x^(27)Cos x + e^(x))dx=
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- int(-pi//2)^(pi//2) Cos x. Sin h x dx=
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- int(0)^(pi) cos^(5) x dx=
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- int(0)^(2pi) sin^(7) x dx=
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- Find the value of int(0)^(pi) (x)/(1+sin x) dx
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- int(0)^(pi) (x Sin x)/(1+Cos^(2)x)dx=
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- If 0 lt a lt b, then int(a)^(b) (|x|)/(x) dx=
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- underset(0)overset(2)int |1-x| dx
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- int(0)^(pi) sqrt((1+cos 2x)/(2))dx=
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- int(0)^(7) [x]dx=
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- int(-1)^(1) (x-[x])dx=
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- int(0)^(2pi) sqrt(1-cos 2x)/(2)dx=
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- int(a)^(b) (|x-a|+|x-b|)dx(0 lt a lt b)=
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- Evaluate underset(-pi//2)overset(pi/2)int sin|x| dx
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- int(0)^(pi//3) |tan x - 1|dx=
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- If f(x) =int(-1)^(x) |t|dt then for any x ge 0, f(x)=
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- If [x] denotes the greatest intger function then the velue of int(0.5)...
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- Evaluate int(0)^(2)[x^(2)] dx where [ ] denotes the GIF
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