Mathematicsss said:Homework Statement
Integral of 1/sqrt(1+x^2) dx
Homework Equations
sin^2theta`+cos^2theta=1
1+tan^2theta=sec^2theta
The Attempt at a Solution
I plugged x=tant --> dx=sec^2t dt
=> integral of 1/sqrt9(1+tan^2t) sec^2t dt
= integral of t = tan^-1t + C
However, another answer I've seen involves sin, why is that?
I meant t=arctan(x)+CRay Vickson said:Setting ##t = \arctan(t) + C## is wrong.
The integral of 1/sqrt(1+x^2) can be solved using the trigonometric substitution method. Let x = tan(u) and dx = sec^2(u) du. After substitution, the integral becomes ∫ sec(u) du which can be solved using integration by parts or by using the formula ∫ sec(x) dx = ln|sec(x) + tan(x)| + C.
Yes, another method to solve this integral is by using the substitution u = 1+x^2. After substitution, the integral becomes ∫ 1/u^(3/2) du which can be solved by using the power rule of integration.
You can check if the integration is correct by differentiating the result. If the derivative of the solution is equal to the original function, then the integral is correct.
Yes, here is a step-by-step solution for integrating 1/sqrt(1+x^2):
Step 1: Let x = tan(u) and dx = sec^2(u) du
Step 2: Substitute the values in the integral, ∫ 1/sqrt(1+x^2) dx = ∫ 1/sqrt(1+tan^2(u)) sec^2(u) du
Step 3: Simplify the expression to get ∫ sec(u) du
Step 4: Solve the integral using integration by parts or the formula ∫ sec(x) dx = ln|sec(x) + tan(x)| + C
Step 5: Substitute back u = arctan(x) to get the final solution.
Yes, this type of integral is commonly used in physics and engineering, particularly in problems involving motion and forces. It is also used in the calculation of arc length and surface area in calculus.