Integrate Simple: Solved [SOLVED] Simple Integration

[salman213]

[SOLVED] Simple Integration

1. Integrate the following

f(x) = sin x/(cos x )^2





3. Well i don't really want to use subsititution or anything since I am pretty sure this can be done very simply but i don't know why i cannot get it...

sin x /cos ^2 x = sinx / 1- sin^2x

i know ln (cos x)^2 will give me (-2(cosx)*sinx)(1/cos^2x)

so that's not good...

 

[rocomath]

Well i don't really want to use subsititution or anything since I am pretty sure this can be done very simply but i don't know why i cannot get it...

How about u-substitution?

 

[silver-rose]

You must use substitution. Using trig identities will only make it messier.

Hint: get f(x) into the form du / u ^2 .

 

[salman213]

yea i just did it with u as well but that weird, cause this question was in my first section for integral calc (where we basically onyl went through basic rules like power rule, and integrals of other trig functions)

but anyways


let u = cosx

du/dx = -sinx
-du=sinxdx

now i integrate - 1/(u)^2 * du
= integral of -u^-2 *du

= u^-1

= cos^-1 x

= 1/cos x

= sec x


thanksss

 

[VietDao29]

yea i just did it with u as well but that weird, cause this question was in my first section for integral calc (where we basically onyl went through basic rules like power rule, and integrals of other trig functions)

but anyways let u = cosx

du/dx = -sinx
-du=sinxdx

now i integrate - 1/(u)^2 * du
= integral of -u^-2 *du

= u^-1

= cos^-1 x

= 1/cos x

= sec xthanksss

Yup, that seems perfectly correct. Except for a small error, you forgot the Constant of Integration "+ C" at the end of the final result. :)

 

[dynamicsolo]

It happens in this case that there is another way to go about this, although it's not something you'd generally spot at first (you notice it after doing the u-substitution). You can also write
(sin x)/[(cos x)^2] as (1/cos x)·(sin x / cos x) = sec x tan x , which is the derivative of sec x . (This at least serves as a check on your result...)

 

[fermio]

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