∫ csc(x)cot(x) dx
To solve the integral ∫ csc(x)cot(x) dx, we can use a substitution method
To solve the integral ∫ csc(x)cot(x) dx, we can use a substitution method. Let’s start by rewriting the integrand using trigonometric identities.
Recall that csc(x) is the reciprocal of sin(x) and cot(x) is the reciprocal of tan(x). We can rewrite the integrand as:
∫ (1/sin(x))(cos(x)/sin(x)) dx
Multiplying the terms together, we get:
∫ (cos(x) / sin^2(x)) dx
Now, let’s use substitution. We’ll substitute u = sin(x), which means du = cos(x) dx. Rearranging this, we have dx = du / cos(x).
Substituting these values, the integral becomes:
∫ (1/u^2) (du / cos(x))
We can simplify this further:
∫ du / (u^2 * cos(x))
Now, let’s express the integration in terms of u:
∫ du / (u^2 * cos(x))
Since u = sin(x), we can express cos(x) in terms of u:
∫ du / (u^2 * sqrt(1 – u^2))
To solve this integral, we need to express it in terms of a known trigonometric function integral. We can use the substitution u = sin(θ), where du = cos(θ) dθ. Solving for cos(θ), we get:
√(1 – u^2) = √(1 – sin^2(θ)) = √(cos^2(θ)) = cos(θ)
Substituting these values back into the integral, we get:
∫ dθ / sin^2(θ)
Recall that csc(x) is the reciprocal of sin(x). So, we have:
∫ (csc^2(θ)) dθ
Now, we can integrate:
∫ csc^2(θ) dθ = -cot(θ) + C
Substituting back for θ, we have:
-cot(θ) + C = -cot(sin^(-1)(u)) + C
Remember that u = sin(x), so this becomes:
-cot(sin^(-1)(sin(x))) + C
Now, we need to express this in terms of x. Since sin(sin^(-1)(y)) = y, we have:
-cot(sin^(-1)(sin(x))) + C = -cot(x) + C
Therefore, the integral of csc(x)cot(x) dx is -cot(x) + C, where C is the constant of integration.
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