# Math Help - Substitution Integrals Help!

1. ## Substitution Integrals Help!

Hi, I am struggling in Chapter 5, Section 5.
I've been absent due to illness, so I'm trying to teach myself.
I've been doing pretty well on most of the problems, but one in particular has been difficult.

integral sinx/ cos^2x dx

Thank you!

2. Originally Posted by brittneyymariee
Hi, I am struggling in Chapter 5, Section 5.
I've been absent due to illness, so I'm trying to teach myself.
I've been doing pretty well on most of the problems, but one in particular has been difficult.

integral sinx/ cos^2x dx

Thank you!
begin with a substitution of $u = \cos x$

do you see why this works?

can you take it from there?

3. well i just tried doing it a different way...and my answer is

1/2 tanxsecx ^2

is that acceptable or has something gone horribly wrong?

yes? no?

4. Originally Posted by brittneyymariee
yes? no?
correct.

you could have also noticed that the original integrand was $\sec x \tan x$, which is the derivative of $\sec x$

so yes, the answer is $\sec x + C$.

good job

5. hint ...

$\frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}} = \, ?$

6. thanks! when i realized that...it was pretty simple.

i got it!
thanks!

7. Originally Posted by brittneyymariee
Hi, I am struggling in Chapter 5, Section 5.
I've been absent due to illness, so I'm trying to teach myself.
I've been doing pretty well on most of the problems, but one in particular has been difficult.

integral sinx/ cos^2x dx

Thank you!
$\int{\frac{\sin{x}}{\cos^2{x}}\,dx}$.

Let $u = \cos{x}$

Therefore $\frac{du}{dx} = -\sin{x}$.

$\int{\frac{\sin{x}}{\cos^2{x}}\,dx} = -\int{\frac{1}{\cos^2{x}}(-\sin{x})\,dx}$

$= -\int{\frac{1}{u^2}\frac{du}{dx}\,dx}$

$= -\int{u^{-2}\,du}$

$= u^{-1} + C$

$= \frac{1}{\cos{x}} + C$.