Answer to Question #338577 in Calculus for Kayle

Question

Use substitution rule to integrate

1.) Integral x³(x⁴+16)³dx

2.) Integral 3x²(x³+5)⁷dx

3.) Integral x³/√1-x⁴ dx

4.) Integral sin x/(cos x)³ dx

Deadline May 9 2022

Hope i will get full explanation to how you solve this, because i need the understandable, because i am absent in that class, thanks bro.

Answer 1

1)

$\int x^3(x^4+16)^3dx$

$u=x^4+16, du=4x^3dx, x^3dx=\dfrac{1}{4}du$

$\int x^3(x^4+16)^3dx=\dfrac{1}{4}\int u^3du=\dfrac{1}{16}u^4+C$

$=\dfrac{1}{16}(x^4+16)^4+C$

2)

$\int 3x^2(x^3+5)^7dx$

$u=x^3+5, du=3x^2dx$

$\int 3x^2(x^3+5)^7dx=\int u^7du=\dfrac{1}{8}u^8+C$

$=\dfrac{1}{8}(x^3+5)^8+C$

3)

$\int \dfrac{x^3}{\sqrt{1-x^4}}dx$

$u=1-x^4, du=-4x^3dx, x^3dx=-\dfrac{1}{4}du$

$\int \dfrac{x^3}{\sqrt{1-x^4}}dx=-\dfrac{1}{4}\int \dfrac{du}{\sqrt{u}}=-\dfrac{1}{2}\sqrt{u}+C$

$=-\dfrac{1}{2}\sqrt{1-x^4}+C$

4)

$\int\dfrac{\sin x}{(\cos x)^3}dx$

$u=\cos x, du=-\sin xdx$

$\int\dfrac{\sin x}{(\cos x)^3}dx=-\int \dfrac{1}{u^3}du=\dfrac{1}{2u^2}+C$

$=\dfrac{1}{2(\cos x)^2}+C$