In Exercises 39–48, use an appropriate substitution and then a...

Question

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.

∫ √(x) / (1 - x³) dx (Hint: Let u = x³/2)

Accepted Answer

Welcome back everyone. Evaluate the integral square root of x 2 divided by 1 minus x the power of 4 dx, where x is between 0 and 1. For this problem, let's rewrite the integral as follows. Let's distribute the square root, and we're going to have square root of x2, which is going to be x because x is positive. So, in the numerator, we have square root of x2 which becomes x, and in the denominator, we're going to have square root of 1 minus x the power of 4. This is followed by dx. What we're going to do from here is use U substitution and we're going to suppose that U is equal to X2 and remember that the next step is to find the derivative of U with respect to X. Which is the derivative of x 2 and that's 2 x. Now we can show that x multiplied by dx is equal to du divided by 2 or 1/2 du. Which allows us to rewrite the integral in terms of you and the U. So XD X and the numerator becomes 1/2 d u. And now in the denominator we have square root of. 1 minus x to the power of 4, which is x22 or basically u 2. We can now factor out the constant 1/2 and integrate d u divided by square root of 1 minus u 2. Notice that this is now an elementary integral. We get 1/2. And the integral of the u divided by the square root of 1 minus u 2 is inverse of sin of u plus a constant of integration c. Finally, we can replace you with x2 to get 1/2. Inverse of sine of x2 + c, which is our final answer for this problem. Thank you for watching.

In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.∫ √(x) / (1 - x³) dx (Hint: Let u = x³/2)

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In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
∫ √(x) / (1 - x³) dx (Hint: Let u = x³/2)