MHB Integral of $\cos^4(t)$: Solution & Explanation

[karush]

$\tiny\text{Whitman 8.7.12 trig integral} $
$$\int\cos^4\left({t}\right) \ dt
=\frac{3t}{8}
+\frac{\sin\left({2t}\right)}{4 }
+\frac{\sin\left({4t}\right)}{32}
+C$$

Didn't know how to break this up
in that the answer has 3 terms + C

$\tiny\text
{from Surf the Nations math study group} \\
$

 

[MarkFL]

I would write:

$$\cos^4(t)=\cos^2(t)\left(1-\sin^2(t)\right)$$

And then look at tome double-angle identities for sine and cosine. :)

 

[karush]

$$\cos^4(t)=\cos^2(t)\left(1-\sin^2(t)\right)$$

$$\cos^2 \left({t}\right)=\frac{\cos\left({2t}\right)+1}{2} \ \ \ \
\sin^2 \left({t}\right)=\frac{1-\cos\left({2t}\right)}{2}$$
That would give
$$\int \frac{\cos\left({2t}\right)+1}{2}
\left(\frac{1-\cos\left({2t}\right)}{2}\right) \ dt $$
Could we assume $u=2t \ \ \ du=2 dt$ at this point.

 

[Greg]

$$\cos^4(t)=\cos^2(t)\left(1-\sin^2(t)\right)=\left(\dfrac{1+\cos(2t)}{2}\right)^2$$

Expand and integrate.

 

[karush]

$$\cos^4(t)
=\left(\dfrac{1+\cos(2t)}{2}\right)^2
=\frac{\cos^2 \left({2t}\right)}{4 }
+\frac{\cos\left({2t}\right)}{2}
+\frac{1}{4 }$$

So

$\displaystyle
I=\frac{1}{4}\int
\cos^2 \left({2t}\right)
+2\cos\left({2t}\right)+1 \ dt $

$u=2t \ \ \ \frac{1}{2 } du=dt$
$\displaystyle
I=
\frac{1}{8}\left[ \int
\cos^2 \left({u}\right) \ du
+2\int \cos\left({u}\right) \ du
+\int 1 \ du \right]$
So far?

 

[Greg]

We can "eyeball" $\dfrac{\cos(2t)}{2}+\dfrac14$, we get $\dfrac{\sin(2t)}{4}+\dfrac{t}{4}$.

That leaves $\dfrac{\cos^2(2t)}{4}$.

 

[Prove It]

We can "eyeball" $\dfrac{\cos(2t)}{2}+\dfrac14$, we get $\dfrac{\sin(2t)}{4}+\dfrac{t}{4}$.

That leaves $\dfrac{\cos^2(2t)}{4}$.

So use the double angle identity again.

 

[MarkFL]

I would write:

$$\cos^4(t)=\cos^2(t)\left(1-\sin^2(t)\right)$$

And then look at tome double-angle identities for sine and cosine. :)

This is what I had in mind:

$$\cos^4(t)=\cos^2(t)\left(1-\sin^2(t)\right)=\cos^2(t)-\sin^2(t)\cos^2(t)=\frac{\cos(2t)+1}{2}-\frac{\sin^2(2t)}{4}=\frac{1}{2}+\frac{\cos(2t)}{2}-\frac{1-\cos(4t)}{8}=\frac{3}{8}+\frac{\cos(2t)}{2}+\frac{\cos(4t)}{8}$$

Then the result you originally posted easily follows. :)

 

[karush]

OK I can see that would integral to the answer

Some other examples, used power reduction but it didn't seem to be much of an advantage

Learned a lot valuable concepts on this one