Question

Question #f3bd9

Answer 1

`int\ (1-sin^4(2x))/cos^4(2x)\ dx=tan(2x)-x+C`

Explanation:

First, we can use the difference of squares formula:
`(a+b)(a-b)=a^2-b^2`

Applying this to the numerator, we get:
`int\ (1-sin^4(2x))/cos^4(2x)\ dx=int\ ((1+sin^2(2x))(1-sin^2(2x)))/cos^4(2x)\ dx`

Next we can use the Pythagorean identity to turn `1-sin^2(2x)` into `cos^2(2x)`:
`int\ ((1+sin^2(2x))cos^2(2x))/cos^4(2x)\ dx=int\ (1+sin^2(2x))/cos^2(2x)\ dx`

Now we can split the fraction into two:
`int\ (1+sin^2(2x))/cos^2(2x)\ dx=int\ 1/cos^2(2x)\ dx+int\ sin^2(2x)/cos^2(2x)\ dx`

I will call the left one Integral 1 and the right one Integral 2

Integral 1
We can use the following identity to rewrite the integral:
`1/cos(theta)=sec(theta)`

`int\ 1/cos^2(2x)\ dx=int\ sec^2(2x)\ dx`

We can now introduce a quick u-substitution with `u=2x`, and divide by `2` to integrate with respect to `u`:
`1/2int\ sec^2(u)\ du`

This is the familiar derivative of `tan(u)`:
`1/2int\ sec^2(u)\ du=1/2tan(u)+C=1/2tan(2x)+C`

Integral 2
We can first use the following identity to rewrite the integral in terms of `tan`:
`sin(theta)/cos(theta)=tan(theta)`

`int\ sin^2(2x)/cos^2(2x)\ dx=int\ tan^2(2x)\ dx`

We can now use this identity to rewrite the integral:
`tan^2(theta)=sec^2(theta)-1`

This gives:
`int\ tan^2(2x)\ dx=int\ sec^2(2x)-1\ dx`

We already worked out the left part in Integral 1, and the right part is just equal to `x`:
`int\ sec^2(2x)-1\ dx=1/2tan(2x)-x+C`

Completing the original integral
Now that we know Integral 1 and Integral 2, we can plug them into our original integral to get the final answer:
`int\ (1-sin^4(2x))/cos^4(2x)\ dx=1/2tan(2x)+1/2tan(2x)-x+C=`

`=tan(2x)-x+C`