U substitution: int [ -4 sqrt(x) sin(1 + x^(3/2)) ] dx

[scrum]

I'm having trouble with u substitution. ( i'm also using tex for the first time)

Evaluate the indefinite integral
\(\displaystyle \int -4\sqrt{x} sin(1+x^{(3/2)}) dx\)

I said

u = \(\displaystyle 1+x^{3/2}\)
du = \(\displaystyle \sqrt{x}dx\)

Then i said \(\displaystyle -4sin(u)du\)

integrated it and got \(\displaystyle 4cos(u)+C\)

sub in u

and got \(\displaystyle 4cos(x^{3/2}+1)+C\)

which is wrong. The answer in the back of the book is \(\displaystyle (8/3) cos(x^{(3/2)}+1)+C\)

I think it's the -4 that is throwing me off but i don't know how to get it to be 8/3

 

[o_O]

Re: U substitution

You forgot to bring the exponent down when taking the derivative of u = 1 + x^(3/2):

\(\displaystyle u = 1 + x^{\frac{3}{2}}\)

\(\displaystyle du = \frac{3}{2}x^{\frac{1}{2}}dx\)

 

[scrum]

Re: U substitution

i'm getting 6cos(1+x^(3/2))+C now.

 

[o_O]

Re: U substitution

\(\displaystyle du = \frac{3}{2} \sqrt{x} dx\)

\(\displaystyle \frac{2}{3}du = \sqrt{x} dx\)


So ...

\(\displaystyle \int -4 sin(1 + x^{\frac{3}{2})} \underbrace{\sqrt{x}dx}_{\frac{2}{3} du}\)

\(\displaystyle = \int -4 \cdot \frac{2}{3} \: \mbox{...}\)

etc. etc.

 

[skeeter]

Re: U substitution

\(\displaystyle u = 1 + x^{\frac{3}{2}}\)
\(\displaystyle du = \frac{3}{2} x^{\frac{1}{2}}\)

\(\displaystyle \int -4 x^{\frac{1}{2}} \sin(1 +x^{\frac{3}{2}}) dx\)

\(\displaystyle -4 \cdot \frac{2}{3} \int \frac{3}{2} x^{\frac{1}{2}} \sin(1 + x^{\frac{3}{2}}) dx\)

\(\displaystyle -\frac{8}{3} \int \sin{u} du\)

\(\displaystyle \frac{8}{3} \cos{u} + C\)

\(\displaystyle \frac{8}{3} \cos(1 + x^{\frac{3}{2}}) + C\)