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AAKASH SERIES-DEFINITE INTEGRALS-PRACTICE EXERCISE
- int(0)^(1)(x^(3))/((1+x^(2))^(3))dx=
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- int(1)^(32) (dx)/(x^(1//5)sqrt(1+x^(4//5)))=
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- If I=int(0)^(1)cos(2Cot^(-1)""(sqrt (1-x)/(1+x)))dx then
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- int(0)^(3)xsqrt(1+x)dx=
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- Evaluate the following integrals (ix) int(0)^(1)(x)/(1+sqrt(x))dx
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- int(log2)^(t)(dx)/(sqrt(e^(x)-1))=(pi)/(6), then t=
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- int(2)^(3)(2-x)/(sqrt(5x-6-x^2))dx=
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- int(1)^(3)(1)/(sqrt(x+1)-sqrt(x-1))dx=
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- int(0)^(pi//4)(sinx + cos x)/(7+9 sin 2x) dx =
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- int(0)^(pi//2) (1)/(9 cos x + 12 sin x) dx=
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- Evaluate int(0)^(pi//2) (1)/(sin x + cos x) dx
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- int(0)^(pi//2) (dx)/(1 + sin^(2)x)=
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- int(0)^(pi) (dx)/(1+2 sin^(2)x)=
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- int(0)^(pi//2) x^(2)sin x dx=
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- int(0)^(1) sin^(-1) sqrt(x)dx=
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- int(0)^(1//sqrt2)(Sin^(-1))/((1-x^(2))^(3//2))dx=
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- int(1)^(2)(e^(x)(1+x log x))/(x) dx=
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- int(1)^(e) e^(x)((x-1)/(x^(2)))dx=
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- int(0)^(2pi) e^(x//2) sin (x/2 + pi/4) dx=
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- int(0)^(pi//2) (x sec^(2). x/2 + 2 tan. x/2)dx=
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