int (sinx/cos x^2) dx

[xomandi]

int (sinx/cos x^2) dx

do i have to convert this to a different identity before trying to integrate??

well sin/cos = tan but we have a cos^2 ahhh I am confused!

Please help!

 

[skeeter]

hint ...

\(\displaystyle \L \frac{\sin{x}}{\cos^2{x}} = \frac{1}{\cos{x}} \cdot \frac{\sin{x}}{\cos{x}}\)

 

[xomandi]

ohh alright
so

1/cos x * sin x/cos x = sec x tan x

int (sec x tan x) dx = sec x + c

is that right?!

 

[mindy88]

couldn't you just do u subsitution?

so, u= cos x
du= sin x

so you would jus have to integrate 1/(u^2)

 

[xomandi]

couldn't you just do u subsitution?

so, u= cos x
du= sin x

so you would jus have to integrate 1/(u^2)

i think i was taught a completely different way
i have no idea what you mean

 

[skeeter]

ohh alright
so

1/cos x * sin x/cos x = sec x tan x

int (sec x tan x) dx = sec x + c

is that right?!

that is correct.

you'll learn the method of substitution soon enough.