Antiderivative of f(x) = arccos 3x

This problem asks to find the antiderivative F(x) of the function f(x) = arccos 3x satisfying the condition F(0) = 0.

Problem

$f (x) = arccos 3 x, F (0) = 0$

Solution

Step 1

The integrand arccos 3x is an inverse trigonometric function. Such integrals are evaluated by integration by parts.

\int u d v = u v - \int v d u

Step 2

Split the integrand arccos 3x dx into two parts: u = arccos 3x and dv = dx. Then du = −3/√(1−9x²) dx and v = x.

u = arccos 3 x, d u = - \frac{3}{\sqrt{1 - 9 x^{2}}} d x

d v = d x, v = x

Step 3

Substitute into the integration by parts formula. The minus sign in du gives a plus before the remaining integral:

\int arccos 3 x d x = x \cdot arccos 3 x + \int \frac{3 x}{\sqrt{1 - 9 x^{2}}} d x

Step 4

To evaluate the remaining integral, bring the factor 3x under the differential sign. Note that d(1−9x²) = −18x dx, so 3x dx = −(1/6) d(1−9x²). Then the integral reduces to a tabular one:

\int \frac{3 x}{\sqrt{1 - 9 x^{2}}} d x = - \frac{1}{6} \int \frac{d (1 - 9 x^{2})}{\sqrt{1 - 9 x^{2}}} = - \frac{1}{3} \sqrt{1 - 9 x^{2}}

Step 5

Thus, the general expression for the antiderivative is:

F (x) = x \cdot arccos 3 x - \frac{1}{3} \sqrt{1 - 9 x^{2}} + C

Step 6

Find the constant C from the condition F(0) = 0. At x = 0: arccos 0 = π/2, but x·arccos 3x = 0, and √(1−0) = 1, therefore:

F (0) = 0 - \frac{1}{3} \sqrt{1} + C = 0

Step 7

From this, C = 1/3. The required antiderivative has the form:

Answer

$F (x) = x \cdot arccos 3 x - \frac{1}{3} \sqrt{1 - 9 x^{2}} + \frac{1}{3}$

Frequently asked questions

How do I find the antiderivative of f(x) = arccos 3x?

Apply integration by parts ∫ u dv = uv − ∫ v du. Choose u = arccos 3x (inverse trigonometric function, simplifies upon differentiation) and dv = dx (so v = x).

Why is du = −3/√(1−9x²) dx?

By the chain rule: (arccos u)′ = −u′/√(1−u²). Here u = 3x, so u′ = 3 and (3x)² = 9x². Thus du = −3/√(1−9x²) dx.

How is the integral ∫ 3x/√(1−9x²) dx evaluated?

Note that d(1−9x²) = −18x dx, so 3x dx = −(1/6) d(1−9x²). Then the integral becomes −(1/6) ∫ dt/√t where t = 1−9x². The tabular integral ∫ dt/√t = 2√t. This gives −(1/3)·√(1−9x²).

Why is C = 1/3 in this problem?

From the condition F(0) = 0. At x = 0: 0·arccos 0 = 0, √(1−0) = 1, therefore F(0) = 0 − 1/3 + C = 0, hence C = 1/3.