1. ## integral with cos

Hi,

Can someone help me with the interval 0 to pie/2 cos to the 5thxdx, please?

This is what I did:

cos2x=1/2(1+cos2x)

int. 0 to pie/2 (cos3x)cos2xdx=

1/2 int. (1+cos2x)cos3dx=

Then would I do a u substitution?

u=cos2x
du=-sin2x(2)dx

Thank you

2. $\cos^5 x = \cos^4 x \cos x = (1-\sin^2 x)^2 \cos x$.

Let $t=\sin x$ now continue.

3. Hi,

Can you explain to me how you got this, please?.

I understand how you went to the second step, but I don't understand how you got the third equation.

After that I need to do u substitution, right?

t=sinx
dt=cosxdx

int. 1 to pie/1 (1-sin2x) squared cosx
=integ. -sin2xcosx

Now, I just need to plug in pie/2 and 1, right?

Thank you

4. Originally Posted by chocolatelover
Hi,

Can you explain to me how you got this, please?.

I understand how you went to the second step, but I don't understand how you got the third equation.
recall that $\cos^2 x = 1 - \sin^2 x$

therefore, $\cos^4 x = \left( \cos^2 x \right)^2 = \left( 1 - \sin^2 x \right)^2$

After that I need to do u substitution, right?

t=sinx
dt=cosxdx
yes

int. 1 to pie/1 (1-sin2x) squared cosx
=integ. -sin2xcosx

Now, I just need to plug in pie/2 and 1, right?

Thank you
what are you doing here? if you made the t-substitution, why do you still have x's in the expressions??

5. Would it be (-sin2t)sq. dt?

then do I need to do integration by parts?

6. Originally Posted by chocolatelover
Would it be (-sin2t)sq. dt?

then do I need to do integration by parts?
no integration by parts necessary (i see that you have a problem knowing when to use by parts and when to use regular substitution. we must address this).

anyway, as TPH showed us:

$\int \cos^5 x dx = \int (1-\sin^2 x)^2 \cos x~dx$

We proceed by substitution.

Let $t=\sin x$

$\Rightarrow dt = \cos x ~dx$

So now, we look on our integral. wherever we see $\sin x$, we replace it with $t$. if we see $\cos x ~dx$, we replace it with $dt$

So our integral becomes:

$\int \left( 1 - t^2 \right)^2~dt$

Hopefully, you can take it from here

Originally Posted by chocolatelover

1/2 int. (1+cos2x)cos3dx=

Then would I do a u substitution?

u=cos2x
du=-sin2x(2)dx

Thank you
by the way. please use "^" when you want to show that something is a power. so to type $\cos^5 x$ you should type cos^5(x) or (cos(x))^5 etc.

### Cos3dx formula

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