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# 𝘶-substitution warmup

Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution.
Find each indefinite integral.

# Problem 1

$\int \mathrm{cos}\left({x}^{2}\right)2x\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

# Problem 2

$\int \frac{3{x}^{2}}{\left({x}^{3}+3{\right)}^{2}}\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

# Problem 3

$\int {e}^{4x}\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

# Problem 4

$\int x\cdot \sqrt{\frac{1}{6}{x}^{2}+1}\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

## Want to join the conversation?

• Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)
• 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).
This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.
• =∫1/u^2 ​du shoudn't be = ln(|u^2|)?
• You are reversing the power rule so the answer is -1/u ​+C. However, integral(1/u) =ln(|u|) + C.
• For problem 2, I don't understand why the antiderivative is - 1/u + C, when we have an indefinite integral of 1/u^2 du? Shouldn't it be 1/U + C?
• Apply the reverse power rule. 1/u^2 is just u^(-2). So, this gives [u^(-2+1)]/(-2 + 1) which is u^(-1)/-1 which is -1/u
• in problem 4 why is xdx= 3?
• if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx
• In problem 2, why the negative?

1/u^2 * du
-1/u
(1 vote)
• Reverse power rule.
∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.
• I was given the problem:
 ∫ sin³(x)cos(x)dx = ? + C

I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.
(1 vote)
• sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.

You need type sin^4(x)/4 or alternatively (sin(x))^4/4.
• This is very hard stuff
• how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)
(1 vote)
• ∫ 4x / sqrt(1 - x^4) dx =
2 ∫ 2x / sqrt(1 - (x^2)^2) dx

Let u = x^2, du = 2x dx, then
2 ∫ 2x dx / sqrt(1 - (x^2)^2) =
2 ∫ du / sqrt(1 - u^2) =
2 arcsin(u) + C =
2 arcsin(x^2) + C.

Hope that I helped.