# Thread: Trouble with integrating trig functions

1. ## Trouble with integrating trig functions

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.

2. 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=sinē(x)+cosē(x)
then transform sin(x)/cos(x)=tan(x)
but it's not necessary

3. 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?

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