- #1
teng125
- 416
- 0
may i know how to integ [2x sin (x^2) dx]??the answer is -cos(x^2)
thanx...
thanx...
teng125 said:may i know how to integ [2x sin (x^2) dx]??the answer is -cos(x^2)
thanx...
Haha, you don't sound to sure on that. Does it make sense?teng125 said:oh...okkk...thanx...
The formula for integrating [2x sin (x^2) dx] is ∫ [2x sin (x^2) dx] = -cos(x^2) + C.
To solve [2x sin (x^2) dx] by substitution, let u = x^2 and du = 2x dx. Then the integral becomes ∫ [sin(u) du], which can be solved using the formula ∫ sin(x) dx = -cos(x) + C.
Yes, you can use integration by parts to solve [2x sin (x^2) dx]. Let u = 2x and dv = sin(x^2) dx. Then du = 2dx and v = -cos(x^2). Using the formula for integration by parts, the integral becomes ∫ [2x sin (x^2) dx] = -2x cos(x^2) - ∫ [-2cos(x^2) dx], which can be solved using the formula ∫ cos(x) dx = sin(x) + C.
Yes, there is a shortcut or trick to solving [2x sin (x^2) dx]. You can use the substitution u = x^2 to convert the integral into ∫ [sin(u) du], which can be easily solved using the formula ∫ sin(x) dx = -cos(x) + C.
The limits of integration for solving [2x sin (x^2) dx] depend on the specific problem or context. Generally, the integral is evaluated between two limits, which are represented by the lower and upper bound of the integral symbol. For example, if the integral is written as ∫ab [2x sin (x^2) dx], the limits of integration would be a and b.