How do I find the antiderivative of f(x) = arcsin 2x?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = arcsin 2x (inverse trigonometric function, simplifies upon differentiation) and dv = dx (so v = x).

Why is du = 2/√(1−4x²) dx?

By the chain rule: (arcsin u)′ = u′/√(1−u²). Here u = 2x, so u′ = 2 and (2x)² = 4x². Thus du = 2/√(1−4x²) dx.

How is the integral ∫ 2x/√(1−4x²) dx evaluated?

Note that d(1−4x²) = −8x dx, so 2x dx = −(1/4) d(1−4x²). Then the integral becomes −(1/4) ∫ dt/√t where t = 1−4x². The tabular integral ∫ dt/√t = 2√t. This gives −(1/2)·√(1−4x²).

Why is C = −1/2 in this problem?

From the condition F(0) = 0. At x = 0: arcsin 0 = 0, √(1−0) = 1, therefore F(0) = 0 + 1/2 + C = 0, hence C = −1/2.

Antiderivative of f(x) = arcsin 2x

This problem asks to find the antiderivative F(x) of the function f(x) = arcsin 2x satisfying the condition F(0) = 0.

Problem

$f (x) = arcsin 2 x, F (0) = 0$

Solution

Step 1

The integrand arcsin 2x is an inverse trigonometric function. Such integrals are evaluated by integration by parts.

Step 2

Split the integrand arcsin 2x dx into two parts: u = arcsin 2x and dv = dx. Then du = 2/√(1−4x²) dx and v = x.

Why is the integration by parts method chosen?

The integrand contains an inverse trigonometric function arcsin 2x, which has no tabular antiderivative. Neither substitution nor tabular formulas work here. The method of integration by parts is specifically designed for such integrals: by differentiating arcsin 2x we obtain a simple rational function with a root, whose integral is evaluated elementarily.

Why is u = arcsin 2x and not u = 1?

In integration by parts, u is chosen to be the function that simplifies upon differentiation, and dv to be what is easily integrated. If we take u = arcsin 2x, then du = 2/√(1−4x²) dx — a root appears, but the remaining integral reduces to the tabular ∫ du/√u. If instead u = 1 and dv = arcsin 2x dx, we would first need the antiderivative of arcsin 2x — that is, the same problem. According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), the inverse trigonometric function comes before the constant, so u = arcsin 2x.

Why is dv = dx and not just 1?

The integration by parts formula: ∫ u dv = uv − ∫ v du. Here dv is the differential of the function v, and by definition dv = v′(x) dx. In our case dv = dx, so v = x. If we took "dv = 1" without dx, the product u · dv = arcsin 2x · 1 = arcsin 2x would be without dx and would not be a differential.

u = arcsin 2 x, d u = \frac{2}{\sqrt{1 - 4 x^{2}}} d x

d v = d x, v = x

Step 3

Substitute into the integration by parts formula:

\int arcsin 2 x d x = x \cdot arcsin 2 x - \int \frac{2 x}{\sqrt{1 - 4 x^{2}}} d x

Step 4

To evaluate the remaining integral, bring the factor 2x under the differential sign. Note that d(1−4x²) = −8x dx, so 2x dx = −(1/4) d(1−4x²). Then the integral reduces to a tabular one:

\int \frac{2 x}{\sqrt{1 - 4 x^{2}}} d x = - \frac{1}{4} \int \frac{d (1 - 4 x^{2})}{\sqrt{1 - 4 x^{2}}} = - \frac{1}{2} \sqrt{1 - 4 x^{2}}

Step 5

Thus, the general expression for the antiderivative is:

F (x) = x \cdot arcsin 2 x + \frac{1}{2} \sqrt{1 - 4 x^{2}} + C

Step 6

Find the constant C from the condition F(0) = 0. At x = 0: arcsin 0 = 0, √(1−0) = 1, therefore:

F (0) = 0 + \frac{1}{2} \sqrt{1} + C = 0

Step 7

From this, C = −1/2. The required antiderivative has the form:

Answer

$F (x) = x \cdot arcsin 2 x + \frac{1}{2} \sqrt{1 - 4 x^{2}} - \frac{1}{2}$