1. ## Integration questions

Hi
Three questions i am having trouble integrating:
1) $\int{\frac{cosx}{sinx}} dx$
This is what i have done:
$\int{\frac{1}{tanx}}$
so i got:
$sec^2x + c$

answers says its: $ln(sinx)+c$

Someone tell me why is my one incorrect?

2) $\int{\frac{x+1}{x^2+2x}} dx$
Need help how to start this.

3) $\int{\frac{1}{\sqrt{1-9x^2}}}$
This is what i got: $sin^{-1}3x+c$

And this what the answers says: $\frac{1}{3}sin^{-1}3x+c$

Someone tell me why is my one incorrect?

P.S

2. #1: How did you get $\sec^2 x$ ? Use the substitution: $u = \sin x \ \Rightarrow \ du = \cos x \ dx$ to get a standard integral

#2: Use the substitution $u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx$ to get another standard integral

#3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: $\int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C$

3. Originally Posted by o_O
#1: How did you get $\sec^2 x$ ? Use the substitution: $u = \sin x \ \Rightarrow \ du = \cos x \ dx$ to get a standard integral
ok i'll try that.

Originally Posted by o_O
#2: Use the substitution $u = x^2 + 2x \ \Rightarrow \ du = 2(x+1)dx$ to get another standard integral
i'll try that.

Originally Posted by o_O
#3: How did you do this question? It looks like you simply used a formula. So if you have a table of integrals, you should see that: $\int \frac{1}{\sqrt{1 - (ax)^2} }\ dx \ = \ \frac{1}{a} \arcsin (ax) + C$
ok i understand now.

4. Originally Posted by Paymemoney
3) $\int{\frac{1}{\sqrt{1-9x^2}}}$
This is what i got: $sin^{-1}3x+c$

And this what the answers says: $\frac{1}{3}sin^{-1}3x+c$
If you don't want to use a formula, then you can use trigonometric substitution.

Notice that

$\int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}$

$= \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}$

$= \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}$.

Now make the substitution $x = \frac{1}{3}\sin{\theta}$ so that $dx = \frac{1}{3}\cos{\theta}\,d\theta$.

Also notice that $\sin{\theta} = 3x$ and so $\theta = \arcsin{3x}$.

Then the integral becomes

$= \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1} {3}\cos{\theta}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9}(1 - \sin^2{\theta})}}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq rt{\cos^2{\theta}}}\,d\theta}$

$= \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

$= \frac{1}{3}\int{1\,d\theta}$

$= \frac{1}{3}\theta + C$

$= \frac{1}{3}\arcsin{3x} + C$.

5. Originally Posted by Prove It
If you don't want to use a formula, then you can use trigonometric substitution.

Notice that

$\int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}$

$= \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}$

$= \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}$.

Now make the substitution $x = \frac{1}{3}\sin{\theta}$ so that $dx = \frac{1}{3}\cos{\theta}\,d\theta$.

Also notice that $\sin{\theta} = 3x$ and so $\theta = \arcsin{3x}$.

Then the integral becomes

$= \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1} {3}\cos{\theta}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9}(1 - \sin^2{\theta})}}\,d\theta}$

$= \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq rt{\cos^2{\theta}}}\,d\theta}$

$= \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

$= \frac{1}{3}\int{1\,d\theta}$

$= \frac{1}{3}\theta + C$

$= \frac{1}{3}\arcsin{3x} + C$.
using the u-substitution looks like i longer way of working out it.

6. This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

$u=sinx$

$du=cosx$

$\int{\frac{cosx}{u}}$

$cosx\int{u}$

$=cosx \frac{u^2}{2} +c$

$=\frac{cos(x)u^2}{2}$

substitute $sinx$back into u $

=\frac{cosxsinx^2}{2}$

7. You can't factor a variable term like $\cos{x}$ from an integral. Instead of factoring it from the integral, replace it with $du$ because, as you stated, $du = \cos{x}$. Then, use the integration rule $\int \frac{du}{u} = \ln|u| + C$.

8. Originally Posted by Paymemoney
This is what i done in the second question. Can you tell me where i have gone wrong, note this is my first time doing u-substitution.

$u=sinx$

$du=cosx$

$\int{\frac{cosx}{u}}$

$cosx\int{u}$

$=cosx \frac{u^2}{2} +c$

$=\frac{cos(x)u^2}{2}$

substitute $sinx$back into u $

=\frac{cosxsinx^2}{2}$
For $\int \frac{cos(x)}{sin(x)} dx$ :

substitute $u=sin(x)$ to get $du=cos(x)dx$

the integral becomes: $\int \frac{du}{u}=ln|u|+C=ln|sin(x)|+C$

you should review the substitution method in your book.

T

9. ok thanks everyone for the help i understand how to do it now.

10. Originally Posted by Paymemoney
using the u-substitution looks like i longer way of working out it.
I'm just showing you how the standard integral was found in the first place - i.e. through trigonometric substitution.