Integrate : ∫(cot2x-csc2x)^2dx
recall that cot^2 x +1 = csc^2 x
so
∫(cot^2 x-csc^2 x)^2dx
= ∫(cot^2 x-(cot^2 + 1)^2dx
= ∫ -1 dx
take over
To integrate the given expression ∫(cot^2x - csc^2x)^2dx, we can expand and simplify the expression before integrating.
Let's start by expanding the square term:
(cot^2x - csc^2x)^2 = (cot^2x - 2cotxcscx + csc^2x)(cot^2x - 2cotxcscx + csc^2x).
Now, distribute and simplify:
= cot^4x - 2cot^3x cscx + cot^2x csc^2x
- 2cot^3x cscx + 4cot^2x csc^2x - 2cotx csc^3x
+ cot^2x csc^2x - 2cotx csc^3x + csc^4x
Combining like terms, we have:
= cot^4x - 4cot^3x cscx + 6cot^2x csc^2x - 4cotx csc^3x + csc^4x
Now, we can integrate each term individually:
∫cot^4x dx - 4∫cot^3x cscx dx + 6∫cot^2x csc^2x dx - 4∫cotx csc^3x dx + ∫csc^4x dx
To solve each integral, we can use trigonometric identities.
The integral of cot^4x can be found by using the identity: cot^2x = csc^2x - 1.
Let's simplify the first integral:
∫cot^4x dx = ∫(cot^2x)^2 dx
Since cot^2x = csc^2x - 1, we can rewrite it as:
∫(csc^2x - 1)^2 dx
Expanding and simplifying:
∫(csc^4x - 2csc^2x + 1) dx
= ∫csc^4x dx - 2∫csc^2x dx + ∫dx
To find the integral of csc^4x, we can use the identity: csc^2x = cot^2x + 1.
The integral becomes:
∫csc^4x dx = ∫(cot^2x + 1)^2 dx
= ∫(cot^4x + 2cot^2x + 1) dx
= ∫cot^4x dx + 2∫cot^2x dx + ∫dx
Now, let's look at each integral:
∫cot^4x dx + 2∫cot^2x dx + ∫dx = ∫csc^4x dx - 2∫csc^2x dx + ∫dx
The integral of ∫csc^4x dx cancels out with -∫csc^4x dx, leaving us with:
2∫cot^2x dx = -2∫csc^2x dx
Next, we can find the integral of ∫cot^2x dx.
Using the identity csc^2x = cot^2x + 1, we have:
∫cot^2x dx = ∫(csc^2x - 1) dx
= ∫csc^2x dx - ∫dx
Substituting this back into our equation, we have:
2∫(csc^2x - 1) dx = -2∫csc^2x dx
Simplifying, we now have:
2∫csc^2x dx = -∫dx
The integral of -∫dx is simply -x.
Finally, our overall integral becomes:
= ∫csc^4x dx - 2∫csc^2x dx + ∫dx
= -x + C
Therefore, the answer to the given integral, ∫(cot^2x - csc^2x)^2dx, is -x + C, where C is the constant of integration.