Answer to Question #338577 in Calculus for Kayle

Question #338577

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.

1
Expert's answer
2022-05-10T13:26:38-0400

1)


x3(x4+16)3dx\int x^3(x^4+16)^3dx

u=x4+16,du=4x3dx,x3dx=14duu=x^4+16, du=4x^3dx, x^3dx=\dfrac{1}{4}du


x3(x4+16)3dx=14u3du=116u4+C\int x^3(x^4+16)^3dx=\dfrac{1}{4}\int u^3du=\dfrac{1}{16}u^4+C

=116(x4+16)4+C=\dfrac{1}{16}(x^4+16)^4+C

2)


3x2(x3+5)7dx\int 3x^2(x^3+5)^7dx

u=x3+5,du=3x2dxu=x^3+5, du=3x^2dx


3x2(x3+5)7dx=u7du=18u8+C\int 3x^2(x^3+5)^7dx=\int u^7du=\dfrac{1}{8}u^8+C

=18(x3+5)8+C=\dfrac{1}{8}(x^3+5)^8+C

3)


x31x4dx\int \dfrac{x^3}{\sqrt{1-x^4}}dx

u=1x4,du=4x3dx,x3dx=14duu=1-x^4, du=-4x^3dx, x^3dx=-\dfrac{1}{4}du


x31x4dx=14duu=12u+C\int \dfrac{x^3}{\sqrt{1-x^4}}dx=-\dfrac{1}{4}\int \dfrac{du}{\sqrt{u}}=-\dfrac{1}{2}\sqrt{u}+C

=121x4+C=-\dfrac{1}{2}\sqrt{1-x^4}+C

4)


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

u=cosx,du=sinxdxu=\cos x, du=-\sin xdx


sinx(cosx)3dx=1u3du=12u2+C\int\dfrac{\sin x}{(\cos x)^3}dx=-\int \dfrac{1}{u^3}du=\dfrac{1}{2u^2}+C

=12(cosx)2+C=\dfrac{1}{2(\cos x)^2}+C


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