Question

I need help with these, please simplify

d. Jo 3x+2 dx 4. Group member #4 should submit only these four problems. a. S e3* sin(e3x) dx C. S dx 19-16x8 sec2(In x) b. S. 73 dx 1 1 3х

Answer 1

a)

e3sin(er)dar

let assume that,

31 U=e

du = 3e3r da

we can say that,

3r sin(3r) d.c = du sin(u) 3

1 le3 sin(e3 )dx = cos(u) +

substitute back u= e^3x,

$\int e^{3x}sin(e^{3x})dx=-\frac{1}{3}cos(e^{3x})+C$

b)

$\int \frac{sec^2(ln(x))}{3x}dx$

let assume that,

$u=ln(x)$

$du= \frac{1}{x}dx$

we can say that,

$\int \frac{sec^2(ln(x))}{3x}dx=\int \frac{sec^2(u)}{3}du$

$\int \frac{sec^2(ln(x))}{3x}dx=\frac{tan(u)}{3}+C$

substitute back u= ln(x),

$\int \frac{sec^2(ln(x))}{3x}dx=\frac{tan(ln(x))}{3}+C$

c)

$\int \frac{x^3}{\sqrt{9-16x^8}}dx=\int \frac{x^3}{\sqrt{9-16(x^4)^2}}dx$

let assume that,

$u=x^4$

$du=4x^3dx$

we can say that,

$\int \frac{x^3}{\sqrt{9-16x^8}}dx=\int \frac{1}{\sqrt{9-16u^2}} \frac{du}{4}$

$\int \frac{x^3}{\sqrt{9-16x^8}}dx=\frac{1}{4}\int \frac{1}{\sqrt{3^2-(4u)^2}} du$

$\int \frac{x^3}{\sqrt{9-16x^8}}dx=\frac{1}{4} \cdot \frac{1}{4}sin^{-1}\left ( \frac{4u}{3} \right )+C$

substitute back u= x⁴,

$\int \frac{x^3}{\sqrt{9-16x^8}}dx=\frac{1}{16}sin^{-1}\left ( \frac{4x^4}{3} \right )+C$

d)

$\int_{0}^{1} \frac{1}{3x+2}dx=\left [ \frac{ln|3x+2|}{3} \right ]_{0}^{1}$

$\int_{0}^{1} \frac{1}{3x+2}dx=\frac{1}{3}\left [ ln|3x+2| \right ]_{0}^{1}$

$\int_{0}^{1} \frac{1}{3x+2}dx=\frac{1}{3} [ ln|3(1)+2|- ln|3(0)+2|]$

$\int_{0}^{1} \frac{1}{3x+2}dx=\frac{1}{3} [ ln|5|- ln|2|]$

$\int_{0}^{1} \frac{1}{3x+2}dx=0.30543$