How do I find this integral using weierstrass substitution?
1/9-4((sinx)^2)
To find the integral of the expression 1/9 - 4(sin^2(x)) using the Weierstrass substitution, we need to follow these steps:
1. Start by expressing sin^2(x) in terms of the Weierstrass substitution, which is u = tan(x/2). Recall the trigonometric identity sin^2(x) = (1 - cos(2x))/2.
2. Substitute sin^2(x) in the integral with (1 - cos(2x))/2, which gives us:
∫ (1/9) - 4(1 - cos(2x))/2 dx.
3. Simplify the expression by expanding everything out:
∫ (1/9) - 4/2 + 4cos(2x)/2 dx.
4. Now, separate the integral into three parts:
∫ (1/9) dx - ∫ (4/2) dx + ∫ (4cos(2x)/2) dx.
5. Integrate each part separately:
∫ (1/9) dx = (1/9) x,
∫ (4/2) dx = 4/2 x = 2x,
∫ (4cos(2x)/2) dx = 4/2 ∫ cos(2x) dx = 2 ∫ cos(2x) dx.
6. Apply the integral of cos(2x) by using the formula:
∫ cos(2x) dx = (1/2) sin(2x).
7. Substitute the integrals back into the original expression:
(1/9) x - 2x + 2(1/2) sin(2x) + C,
which simplifies to:
(1/9) x - 2x + sin(2x) + C.
Therefore, the integral of 1/9 - 4(sin^2(x)) using the Weierstrass substitution is (1/9) x - 2x + sin(2x) + C.