Proof
- Parts with u = sin⁻¹x, dv = dx, so du = dx/√(1 − x²), v = x.
- I = x sin⁻¹x − ∫ x/√(1 − x²) dx.
- Put t = 1 − x²: ∫ x/√(1 − x²) dx = −√(1 − x²). So I = x sin⁻¹x + √(1 − x²) + C.
Check it: differentiate the answer and you should get back sin⁻¹x. C is the constant of integration; leave it out only in definite integrals.
Standard integration formulas
| Integral | Result |
|---|---|
| ∫ sin x dx | −cos x + C |
| ∫ cos x dx | sin x + C |
| ∫ tan x dx | log|sec x| + C |
| ∫ cot x dx | log|sin x| + C |
| ∫ sec x dx | log|sec x + tan x| + C |
| ∫ cosec x dx | log|cosec x − cot x| + C |
| ∫ sec²x dx | tan x + C |
| ∫ cosec²x dx | −cot x + C |
| ∫ sec x tan x dx | sec x + C |
| ∫ cosec x cot x dx | −cosec x + C |
| ∫ log x dx | x log x − x + C |
| ∫ eˣ dx | eˣ + C |
| ∫ aˣ dx | aˣ / log a + C |
| ∫ 1/x dx | log|x| + C |
| ∫ xⁿ dx | xⁿ⁺¹/(n + 1) + C, n ≠ −1 |
| ∫ sin²x dx | x/2 − (sin 2x)/4 + C |
| ∫ cos²x dx | x/2 + (sin 2x)/4 + C |
| ∫ tan²x dx | tan x − x + C |
| ∫ sin³x dx | cos³x/3 − cos x + C |
| ∫ sin x cos x dx | (sin²x)/2 + C |
| ∫ sec³x dx | ½(sec x tan x + log|sec x + tan x|) + C |
| ∫ 1/(1 + x²) dx | tan⁻¹x + C |
| ∫ 1/√(1 − x²) dx | sin⁻¹x + C |
| ∫ 1/(x² + a²) dx | (1/a) tan⁻¹(x/a) + C |
| ∫ 1/(x² − a²) dx | (1/2a) log|(x − a)/(x + a)| + C |
| ∫ 1/(a² − x²) dx | (1/2a) log|(a + x)/(a − x)| + C |
| ∫ 1/√(a² − x²) dx | sin⁻¹(x/a) + C |
| ∫ 1/√(x² + a²) dx | log|x + √(x² + a²)| + C |
| ∫ 1/√(x² − a²) dx | log|x + √(x² − a²)| + C |
| ∫ √(a² − x²) dx | (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C |
| ∫ √(x² + a²) dx | (x/2)√(x² + a²) + (a²/2) log|x + √(x² + a²)| + C |
| ∫ √(x² − a²) dx | (x/2)√(x² − a²) − (a²/2) log|x + √(x² − a²)| + C |
| ∫ x eˣ dx | (x − 1)eˣ + C |
| ∫ x sin x dx | sin x − x cos x + C |
| ∫ x cos x dx | x sin x + cos x + C |
| ∫ x log x dx | (x²/2) log x − x²/4 + C |
| ∫ eˣ sin x dx | (eˣ/2)(sin x − cos x) + C |
| ∫ eˣ cos x dx | (eˣ/2)(sin x + cos x) + C |
| ∫ sin⁻¹x dx | x sin⁻¹x + √(1 − x²) + C |
| ∫ tan⁻¹x dx | x tan⁻¹x − ½ log(1 + x²) + C |
| ∫ 1/(1 + cos x) dx | tan(x/2) + C |
| ∫ 1/(1 + sin x) dx | tan x − sec x + C |
| ∫ (log x)/x dx | (log x)²/2 + C |
| ∫ 1/(x log x) dx | log|log x| + C |
| ∫ eˣ[f(x) + f′(x)] dx | eˣ f(x) + C |
Questions students ask
What is the integration of sin⁻¹x?
∫ sin⁻¹x dx = x sin⁻¹x + √(1 − x²) + C.
How do you integrate sin⁻¹x?
Parts with u = sin⁻¹x, dv = dx, so du = dx/√(1 − x²), v = x. I = x sin⁻¹x − ∫ x/√(1 − x²) dx. Put t = 1 − x²: ∫ x/√(1 − x²) dx = −√(1 − x²). So I = x sin⁻¹x + √(1 − x²) + C.
Which method is used to integrate sin⁻¹x?
By parts.