Find the integral with respect to x and simplify

Find the integral with respect to x and simplify: a) \int \sin2x\cos4x\sin4xdx [3] b) \int \frac{2}{x^2\sqrt{1+x^2}}dx [5] c) \int \frac{4x+2}{\sqrt{x^2-16}}dx [5] d) \int x^3e^{2x} dx [5] e) \int \sqrt{x}\ln2x dx [5] f) \int \sin^{-1}2x dx [5] g) \int \cos^43x dx [5] h) \int \frac{3x^2+16x+15}{(x+3)^3}dx [5] i) \int \frac{3+6x+4x^2-2x^3}{(x^4+3x^2)}dx [5] j) \int \frac{x^4-7x-2}{x^3-8}dx [7]

The Correct Answer and Explanation is:

Certainly, Jeniffer. Let’s tackle these integrals carefully, using the appropriate methods for each. Due to their complexity and your request for a 300-word explanation, I’ll summarize the techniques used, then provide the solutions.

Overview of Techniques

The integrals involve a mix of strategies: trigonometric identities, u-substitution, integration by parts, partial fraction decomposition, and trigonometric powers reduction. Let’s break down the more intricate ones and outline key ideas, focusing on variety and precision over brevity.

Selected Solutions and Explanation (abridged):

a) ∫sin⁡2xcos⁡4xsin⁡4x dx\int \sin 2x \cos 4x \sin 4x \, dx

Let’s simplify cos⁡4xsin⁡4x=12sin⁡8x\cos 4x \sin 4x = \frac{1}{2} \sin 8x. Then:

∫sin⁡2x⋅12sin⁡8x dx=12∫sin⁡2xsin⁡8x dx\int \sin 2x \cdot \frac{1}{2} \sin 8x \, dx = \frac{1}{2} \int \sin 2x \sin 8x \, dx

Use product-to-sum:

sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A \sin B = \frac{1}{2}[\cos(A-B) – \cos(A+B)]

⇒14[∫cos⁡(6x)−cos⁡(10x) dx]=14(sin⁡6×6−sin⁡10×10)+C\Rightarrow \frac{1}{4} \left[ \int \cos(6x) – \cos(10x) \, dx \right] = \frac{1}{4} \left( \frac{\sin 6x}{6} – \frac{\sin 10x}{10} \right) + C

b) ∫2×21+x2 dx\int \frac{2}{x^2 \sqrt{1+x^2}} \, dx

Let x=tan⁡θ⇒dx=sec⁡2θdθx = \tan \theta \Rightarrow dx = \sec^2 \theta d\theta, then it becomes:

∫2tan⁡2θ⋅sec⁡θ⋅sec⁡2θdθ\int \frac{2}{\tan^2 \theta \cdot \sec \theta} \cdot \sec^2 \theta d\theta

Simplify and back-substitute to get:

−2×1+x2+C\boxed{-\frac{2}{x \sqrt{1+x^2}} + C}

d) ∫x3e2x dx\int x^3 e^{2x} \, dx

Use integration by parts repeatedly or tabular method. Final result:

e2x(x32−3×24+3×4−38)+C\boxed{e^{2x} \left( \frac{x^3}{2} – \frac{3x^2}{4} + \frac{3x}{4} – \frac{3}{8} \right) + C}

The key to mastering integrals lies in recognizing structure and applying substitutions or expansions strategically. If you want full solutions for the remaining ones, I can walk through them too. Shall we continue?