Antiderivative Of Arccos: The Step Most People Miss
The antiderivative of arccos(x) is $$x\,\arccos(x)-\sqrt{1-x^2}+C$$, and the key step most people miss is using integration by parts with $$u=\arccos(x)$$ and $$dv=dx$$. The standard result is consistent across calculus references and step-by-step worked solutions.
Why this integral matters
The integral $$\int \arccos(x)\,dx$$ is a classic inverse-trig problem because it does not simplify by substitution alone; it requires a deliberate choice of parts. That is why many students can recognize the function but still miss the correct setup on the first attempt.
In practical terms, the solution is a good test of whether you can connect inverse-trigonometric differentiation with integration technique, which is a foundational skill in calculus.
Step-by-step derivation
- Set $$u=\arccos(x)$$ and $$dv=dx$$.
- Then $$du=-\frac{1}{\sqrt{1-x^2}}\,dx$$ and $$v=x$$.
- Apply integration by parts: $$\int u\,dv = uv-\int v\,du$$.
- Substitute the pieces: $$\int \arccos(x)\,dx = x\arccos(x)+\int \frac{x}{\sqrt{1-x^2}}\,dx$$.
- Use the substitution $$t=1-x^2$$, so $$dt=-2x\,dx$$.
- That gives $$\int \frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}$$.
- Combine terms to get $$\int \arccos(x)\,dx = x\arccos(x)-\sqrt{1-x^2}+C$$.
Core formula
| Expression | Result | Notes |
|---|---|---|
| $$\int \arccos(x)\,dx$$ | $$x\arccos(x)-\sqrt{1-x^2}+C$$ | Valid for $$-1<x<1$$. |
| $$\frac{d}{dx}\arccos(x)$$ | $$-\frac{1}{\sqrt{1-x^2}}$$ | This derivative drives the integration-by-parts setup. |
| $$\int \frac{x}{\sqrt{1-x^2}}\,dx$$ | $$-\sqrt{1-x^2}+C$$ | Usually handled with a direct substitution. |
Where students slip
The most common mistake is choosing $$dv=\arccos(x)\,dx$$, which makes the integral harder instead of easier. The better move is to treat $$\arccos(x)$$ as the differentiable part and $$dx$$ as the easy-to-integrate part.
Another frequent error is forgetting that the derivative of $$\arccos(x)$$ is negative, which changes the sign of the final answer. That sign is not cosmetic; it is the reason the result becomes $$x\arccos(x)-\sqrt{1-x^2}+C$$, not a plus sign in front of the radical.
Quick verification
Differentiating $$x\arccos(x)-\sqrt{1-x^2}$$ returns $$\arccos(x)$$, because the product rule and chain rule cancel the extra terms. This confirmation is a reliable way to check the antiderivative before using it in homework, exams, or applied calculus problems.
"Integration by parts is the decisive move." That is the practical lesson behind this integral, because the inverse cosine does not yield cleanly to a simple substitution on its own.
Student-facing takeaway
If you remember only one thing, remember this: treat $$\arccos(x)$$ as the part you differentiate, not the part you integrate first. That single decision turns a stuck problem into a standard textbook solution.
- $$\int \arccos(x)\,dx = x\arccos(x)-\sqrt{1-x^2}+C$$.
- The essential method is integration by parts.
- The derivative $$\frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1-x^2}}$$ determines the sign.
- A final derivative check is the fastest way to verify the result.
Everything you need to know about Antiderivative Of Arccos The Step Most People Miss
What is the antiderivative of arccos(x)?
The antiderivative of $$\arccos(x)$$ is $$x\arccos(x)-\sqrt{1-x^2}+C$$. This is the standard closed-form result taught in calculus references and worked examples.
Why use integration by parts?
Because $$\arccos(x)$$ becomes simpler when differentiated, while $$dx$$ integrates immediately to $$x$$. That pairing is exactly why integration by parts is the correct method.
How do I check the answer?
Differentiate $$x\arccos(x)-\sqrt{1-x^2}$$ and simplify. If the result is $$\arccos(x)$$, the antiderivative is correct.
