For a general case of the integral of (x^m)(a + bx^n)^p where m, n and p are rational numbers and a and b are real numbers: 1) if p is an integer then substitute x = t^(LCM of the denominators of m and n); 2) if (m + 1)/n is integer then substitute a + bx^n = t^q where q is the denominator of p; 3) if (m + 1)/n + p is an integer then substitute b + a/x^n = t^q where q is tge denominator of p. So after rewriting this integral in the binomial expression we get the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx. m = 3, n = 2, p = -3/2. 1) p is not an integer 2) (m + 1)/n = (3 + 1)/2 = 2 is an integer so 9 + 4x^2 = t^2 4xdx = tdt xdx = tdt/4 x^2 = (t^2 - 9)/4 x^3dx = t(t^2 - 9)dt/16 x = 0 -> t = 3, x = 3sqrt(3)/2 -> t = 6. Finally, the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx is equal to the integral from 3 to 6 of t(t^2 - 9)dt/(16t^3) = (1 - 9t^(-2))dt/16 = (t + 9/t)/16 where t goes from 3 to 6 so it equals to (6 + 9/6 - 3 - 9/3)/16 = 3/32.
Sir, I want to ask that condition will I find the value of the limits, or is it every time I have definite integrals, I will take their new value of u. Coz, I don't understand that. I just want to know the condition behind it. 🙇
We can factor from numerator derivative of the inside of denominator and we are left with x^2 whch can be expressed as function of sqrt(4x^2+9) so change of variable u=sqrt(4x^2+9) is good idea
I get the same result without the x^3=x^2 * x trick. At one point I get [sqrt(u-9)]^3 in the numerator and sqrt(u-9) in the denominator. After cancellation left with [sqrt(u-9)]^2 which conveniently is just u-9.
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For a general case of the integral of (x^m)(a + bx^n)^p where m, n and p are rational numbers and a and b are real numbers:
1) if p is an integer then substitute x = t^(LCM of the denominators of m and n);
2) if (m + 1)/n is integer then substitute a + bx^n = t^q where q is the denominator of p;
3) if (m + 1)/n + p is an integer then substitute b + a/x^n = t^q where q is tge denominator of p.
So after rewriting this integral in the binomial expression we get the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx.
m = 3, n = 2, p = -3/2.
1) p is not an integer
2) (m + 1)/n = (3 + 1)/2 = 2 is an integer so 9 + 4x^2 = t^2
4xdx = tdt
xdx = tdt/4
x^2 = (t^2 - 9)/4
x^3dx = t(t^2 - 9)dt/16
x = 0 -> t = 3, x = 3sqrt(3)/2 -> t = 6.
Finally, the integral from 0 to 3sqrt(3)/2 of (x^3)(9 + 4x^2)^(3/2)dx is equal to the integral from 3 to 6 of t(t^2 - 9)dt/(16t^3) = (1 - 9t^(-2))dt/16 = (t + 9/t)/16 where t goes from 3 to 6 so it equals to (6 + 9/6 - 3 - 9/3)/16 = 3/32.
Nice prof
Sir, I want to ask that condition will I find the value of the limits, or is it every time I have definite integrals, I will take their new value of u.
Coz, I don't understand that.
I just want to know the condition behind it.
🙇
You could use integration by parts method
Yeah.. better to know them both as the question can be direct sometimes 😊
We can factor from numerator derivative of the inside of denominator and we are left with x^2 whch can be expressed as function of sqrt(4x^2+9)
so change of variable u=sqrt(4x^2+9) is good idea
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I get the same result without the x^3=x^2 * x trick. At one point I get [sqrt(u-9)]^3 in the numerator and sqrt(u-9) in the denominator. After cancellation left with [sqrt(u-9)]^2 which conveniently is just u-9.
Its just 1st step game
Aftee u know u have break x³ to x² and x.
After it its easy
Thank you Ifrom Iraq f
Find d²y/dx² when y=square root of x/(x²+x)
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