Trouble with integrating trig functions

**fogel1497** · September 13th 2008, 01:35 PM

I have two trig functions that i cannot seem to find a way to integrate. They are:

INTEGRAL of: 31(1+cos(x))^2
This one looks pretty simple but whenever i check my answer my taking the derivative of it, it never comes out right.

and

INTEGRAL of: 27(sin(x)/(cos(x))^3
Im suppose to have my answer in terms of tangent. Which messes me up because usually i would do a trig substitution to cancel out the sine. If its not possible to express it in terms of tangent then what is it in terms of sine and cosine.

**Moo** · September 13th 2008, 01:46 PM

Hello,

Originally Posted by fogel1497

I have two trig functions that i cannot seem to find a way to integrate. They are:

INTEGRAL of: 31(1+cos(x))^2
This one looks pretty simple but whenever i check my answer my taking the derivative of it, it never comes out right.

$(1+\cos(x))^2=\underbrace{1+2 \cos(x)}_{\text{easy to integrate}}+\cos^2(x)$

Now note that $\cos^2(x)=\frac{1+\cos(2x)}{2}$

and

INTEGRAL of: 27(sin(x)/(cos(x))^3
Im suppose to have my answer in terms of tangent. Which messes me up because usually i would do a trig substitution to cancel out the sine. If its not possible to express it in terms of tangent then what is it in terms of sine and cosine.

Substitute u=cos(x)

get your answer and it'll be ok

if you want to transform into tangent, use the identity 1=sin²(x)+cos²(x)
then transform sin(x)/cos(x)=tan(x)
but it's not necessary

**Soroban** · September 13th 2008, 02:02 PM

Hello, fogel1497!

$\int \frac{27\sin x}{\cos^3\!x}\,dx$

We have: . $27\int\frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2\!x}\,dx \;=\;27\int \tan x\sec^2\!x\,dx$

Now let: $u \:= \:\tan x\quad\Rightarrow\quad du \:= \:\sec^2\!x\,dx$

Got it?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could get a different (but equivalent) answer . . .

We have: . $27\int \sec x \,(\sec x\tan x\,dx)$

Let: $u \:=\:\sec x \quad\Rightarrow\quad du \:=\:\sec x\tan x\,dx$

Substitute: . $27\int u\,du \:=\;\frac{27}{2}u^2 + C$

Back-substitute: . $\frac{27}{2}\sec^2\!x + C$

Thread: Trouble with integrating trig functions

LinkBack

Thread Tools

Display

Trouble with integrating trig functions

Similar Math Help Forum Discussions

Integrating trig functions

Integrating trig functions

Help Integrating Trig functions

[SOLVED] Integrating trig functions

need some calrification on integrating trig functions

Search Tags