Question

when we take antiderivatives. it is look at the integrand and think about taking the derivative. rule, then the antiderivative will require u - substitution.

u - substitution.

\\(\int 6x^2(x^3 + 4)^5 dx\\)

1. \\(\int \frac{(\sqrt{x} - 1)^2}{\sqrt{x}} dx\\)

1. \\(\int \cot(3x) dx\\)

1. \\(\int \frac{1}{\sqrt{1 - 4x^2}} dx\\)

inverse trig can be confused with u - sub.

solve for x.

Explanation:

To solve the integral \(\int \frac{1}{\sqrt{1 - 4x^2}} \, dx\) (problem 7), we use the inverse trigonometric substitution formula for \(\int \frac{1}{\sqrt{1 - u^2}} \, du=\sin^{-1}(u)+C\).

Step 1: Rewrite the Integrand

Rewrite the denominator to match the form \(\sqrt{1 - u^2}\):
\[
\int \frac{1}{\sqrt{1 - (2x)^2}} \, dx
\]

Step 2: Apply \(u\)-Substitution

Let \(u = 2x\). Then, differentiate \(u\) with respect to \(x\):
\[
\frac{du}{dx}=2 \implies dx=\frac{du}{2}
\]

Step 3: Substitute \(u\) and \(dx\)

Substitute \(u = 2x\) and \(dx=\frac{du}{2}\) into the integral:
\[
\int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{2}
\]

Step 4: Simplify and Integrate

Factor out the constant \(\frac{1}{2}\) and integrate using the inverse sine formula:
\[
\frac{1}{2} \int \frac{1}{\sqrt{1 - u^2}} \, du=\frac{1}{2} \sin^{-1}(u)+C
\]

Step 5: Substitute Back \(u\)

Replace \(u\) with \(2x\) to get the final answer:
\[
\frac{1}{2} \sin^{-1}(2x)+C
\]

Answer:

\(\boxed{\frac{1}{2} \sin^{-1}(2x) + C}\)