1. ## More integrals

In the attached picture for #5 what would I use as u, since no term from the problem can be divided out of the original function...

And for #6 I have no idea what I would use as u either. Is there a method I can use to determine this easily in problems like this, because it seems like when people give me the first and second step I can do the rest easily.

2. Hello, fattydq!

Exactly where is your difficulty with #5?

$5)\;\;\int\cot(15x)\,dx$

There is a standard integration formula: . $\int \cot u\,du \;=\;\ln|\sin u| + C$

And obviously, we let: $u = 15x$

3. Originally Posted by fattydq
In the attached picture for #5 what would I use as u, since no term from the problem can be divided out of the original function...

And for #6 I have no idea what I would use as u either. Is there a method I can use to determine this easily in problems like this, because it seems like when people give me the first and second step I can do the rest easily.
for Q5 just use $u = 15x$, then

$\int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du$

for Q6 I would write the integrand as

$\begin{array}{rcl}
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}$

Hope this helps.

4. Originally Posted by Rincewind
for Q5 just use $u = 15x$, then

$\int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du$

for Q6 I would write the integrand as

$\begin{array}{rcl}
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}$

Hope this helps.
OK but how would I figure out the antiderivative of cot?

5. Originally Posted by fattydq
OK but how would I figure out the antiderivative of cot?
It is is many tables of standard integrals. To do it from first principles.

$I = \int\cot(u)du = \int\frac{\cos(u)}{\sin(u)}du$

let $t = \sin(u)$, therefore $dt/du = \cos(u)$ so

$I = \int\frac{dt}{t} = \ln|t| + C = \ln|\sin u| + C$

Hope this helps.

6. Originally Posted by Rincewind
for Q5 just use $u = 15x$, then

$\int\cot(15x)\,dx = \frac{1}{15}\int\cot(u)\,du$

for Q6 I would write the integrand as

$\begin{array}{rcl}
\dfrac{x}{(x+16)^{1/4}} &=& \dfrac{x+16-16}{(x+16)^{1/4}} \\
&=& (x+16)^{3/4} - \dfrac{16}{(x+16)^{1/4}}\end{array}$

Hope this helps.
OK I understand that I'd use (x+16)^(3/4) as u and thus du would be 3/4(x+16)^1/4 and I'd multiply it by 4/3 to even it out so I'm left with 4/3 the integral of -16, 4 times 16 is 64 so my final answer is -64/3 but it's incorrect apparently...where am I going wrong?

7. Originally Posted by fattydq
OK I understand that I'd use (x+16)^(3/4) as u and thus du would be 3/4(x+16)^1/4 and I'd multiply it by 4/3 to even it out so I'm left with 4/3 the integral of -16, 4 times 16 is 64 so my final answer is -64/3 but it's incorrect apparently...where am I going wrong?
I'd just use $u = (x+16)$

$\begin{array}{rcl}\int\dfrac{x\,dx}{\sqrt[4]{x+16}} &=& \int u^{3/4} - 16u^{-1/4}\,du\\
&=& \dfrac{4}{7}u^{7/4} - \dfrac{64}{3}u^{3/4} + C.\end{array}$