# Thread: Find the Indicated Integrals

1. ## Find the Indicated Integrals

I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

The three are:

$\displaystyle \int ln(x^4)/x$ dx

$\displaystyle \int$((e^t^x) cos (e^t))/(3+5 sin (e^t))

and lastly

$\displaystyle \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$

2. Hi
Originally Posted by Pikeman85
$\displaystyle \int ln(x^4)/x$ dx
$\displaystyle \frac{\ln(x^4)}{x}=\frac{4\ln x}{x}=4\times \frac{1}{x}\times \ln x$ and as the derivative of $\displaystyle x\mapsto \ln x$ is $\displaystyle x\mapsto\frac{1}{x}$...
$\displaystyle \int$((e^t^x) cos (e^t))/(3+5 sin (e^t))
Is it $\displaystyle \int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}x}$ or $\displaystyle \int \frac{\mathrm{e}^{tx} \cos (\mathrm{e}^t)}{3+5 \sin (\mathrm{e}^t)}\,\mathrm{d}{\color{red}t}$ ?

$\displaystyle \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$
$\displaystyle \int_0^{\frac{4}{5}} \frac{\arcsin \left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,\mathrm{d}x$
Remember that $\displaystyle \arcsin'x=\frac{1}{\sqrt{1-x^2}}$

3. Originally Posted by Pikeman85
I'm supposed to find several indefinite integrals, but I am not sure exactly how to do each, especially as they involve natural logs and trigonmetric functions.

The three are:

$\displaystyle \int ln(x^4)/x$ dx

$\displaystyle \int$((e^t^x) cos (e^t))/(3+5 sin (e^t))

and lastly

$\displaystyle \int (superscript 4/5, subscript 0) (sin^-1 ((5/4)x))/(sqrt(16-25x^2))$

for the frist one

$\displaystyle \int \frac{\ln(x^4)}{x}dx=4\int\frac{\ln(x)}{x}dx$

Let $\displaystyle u=\ln(x) \implies du=\frac{1}{x}dx$

$\displaystyle 4\int udu=2u^2+C=2\left( \ln(x)\right)^2+C$

For number 2 you have both x's and t's but no dx or dt what variable are integrating with respect to?

For the last one

$\displaystyle \int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx$

let $\displaystyle x=\frac{4}{5}\sin(t) \implies dx=\frac{4}{5}\cos(t)dt$

$\displaystyle \int_{0}^{4/5}\frac{\sin^{-1\left( \frac{5}{4}x\right)}}{\sqrt{16-25x^2}}dx=\int_{0}^{\frac{\pi}{2}}\frac{t}{\sqrt{1 6-16\sin^2(t)}}\left( \frac{4}{5}\cos(t)dt \right)=$

$\displaystyle \frac{1}{5}\int_{0}^{\pi/2}tdt=\frac{1}{5} \left[ \frac{1}{2} \left( \frac{\pi}{2}\right)^2-0\right]=\frac{\pi^2}{40}$

Here is a web page with some La tex code for you
Helpisplaying a formula - Wikipedia, the free encyclopedia
Good luck.

4. Hello, Pikeman85!

These all require "simple" substitutions.
. . The trick is to recognize them.

$\displaystyle \int \frac{\ln(x^4)}{x}\,dx$
We have: .$\displaystyle \int\frac{4\ln(x)}{x}\,dx \;=\;4\int \ln x\,\frac{dx}{x}$

Let $\displaystyle u \:=\:\ln(x) \quad\Rightarrow\quad du \:=\:\frac{dx}{x}$

Substitute: .$\displaystyle 4\int u\,du$ . . . etc.

$\displaystyle \int \frac{e^{t}\cos(e^t)\,dt}{3+5\sin(e^t)}$
Let $\displaystyle u \:=\:3+5\sin(e^t)\quad\Rightarrow\quad du \:=\:5e^t\cos(e^t)\,dt \quad\Rightarrow\quad e^t\cos(e^t)\,dt \:=\:\frac{1}{5}\,du$

Substitute: .$\displaystyle \int\frac{\frac{1}{5}\,du}{u} \;=\;\frac{1}{5}\int \frac{du}{u}$ . . . etc.

$\displaystyle \int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)}{\sqrt{16-25x^2}}\,dx$
The denominator is: .$\displaystyle \sqrt{16\left(1 - \frac{25}{16}x^2\right)} \;=\;4\sqrt{1 - \left(\frac{5}{4}x\right)^2}$

The integral becomes: .$\displaystyle \int^{\frac{4}{5}}_0 \frac{\sin^{-1}\!\left(\frac{5}{4}x\right)\,dx} {4\sqrt{1 - \left(\frac{5}{4}x\right)^2}}$ . $\displaystyle = \;\;\frac{1}{4}\int^{\frac{4}{5}}_0\sin^{-1}\!\left(\frac{5}{4}x\right)\cdot\frac{dx}{\sqrt{ 1 - \left(\frac{5}{4}x\right)^2}}$

$\displaystyle \text{Let }u \:=\:\sin^{-1}\!\left(\frac{5}{4}x\right) \quad\Rightarrow\quad du \:=\:\frac{\frac{5}{4}\,dx}{\sqrt{1-\left(\frac{5}{4}x\right)^2}} \quad\Rightarrow\quad \frac{dx}{\sqrt{1-\left(\frac{5}{4}x\right)^2}} \:=\:\frac{4}{5}\,du$

Substitute: .$\displaystyle \frac{1}{4}\int^{\frac{4}{5}}_0 u\cdot\frac{4}{5}\,du \;=\;\frac{1}{5}\int^{\frac{4}{5}}_0 u\,du$ . . . etc.

Edit: I'm way too slow this time . . . *sigh*
.

$\displaystyle \int x^2 \sqrt{(7 + x^3)}$ dx

The answer I got for it is $\displaystyle (x^3/3) \sqrt{(7x+(x^4/4))}$

Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.

6. Originally Posted by Pikeman85

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

The answer I got for it is (x^3/3) sqrt(7x+(x^4/4))

Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
One can notice that the derivative of $\displaystyle 7+x^3$ is $\displaystyle 3x^2$ so $\displaystyle x^2 \sqrt{7 + x^3}$ looks like $\displaystyle u'(x)\sqrt{u(x)}$ which can easily be integrated : you may try to substitute $\displaystyle u(x)=x^3+7$.

7. Well, a faster substitution is also $\displaystyle z^2=x^3+7.$

8. $\displaystyle x^3 \sqrt{7+x^3}$

I got this, but it did not work. I'm missing a step here I think

I don't get substitution very well

9. Originally Posted by Pikeman85

$\displaystyle \int x^2 \sqrt{(7 + x^3)}$ dx

The answer I got for it is $\displaystyle (x^3/3) \sqrt{(7x+(x^4/4))}$

Which is not correct. How am I doing this incorrectly? I imagine my substitution is wrong.
Let $\displaystyle u=7+x^3 \implies du=3x^2dx \iff \frac{du}{3}=x^2dx$

Now subbing these into your integral gives

$\displaystyle \int x^2\sqrt{7+x^3}dx=\int \sqrt{u}\left( \frac{du}{3}\right)=\frac{1}{3}\int u^\frac{1}{2}du=\frac{2}{9}u^\frac{3}{2}+C=\frac{2 }{9}(7+x^3)^\frac{3}{2}+C$