Evaluate the integral

Question

∫(from 0 to √(3)/4) dx/(1+16x²)

Answer 1

You can recognize 1/(1+u²) as the derivative tan^-1(u) and use the substitution u=4x

You can also use the trig substitution x = (1/4)tan(u) which derives the inverse tangent of x as the solution.

The first sub gives the 1/4 (tan^-1(u)) evaluated from 0 to sqrt(3) = π/12

The second sub approach leads to the integral of 1/(1+tan²(u)) * 1/4 * sec²(u)du = the integral of 1/4 du = 1/4 u or

1/4 ^ tan^-1(4x) from 0 to sqrt(3)/4 (same result)

Please consider a tutor. Take care.

Answer 2

∫_{(0 to √3/4)} [1 / (1+16x²)]dx

Let u = 4x. Then du = 4dx. So, dx = (1/4)du. When x = 0, u = 0 and when x = √3/4, u = √3.

So, we have (1/4)∫_{(0 to √3)} [1 / (1+u²)]du = (1/4)Arctanu_{(0 to √3)}

(1/4)[Arctan√3 - Arctan0] = (1/4)[π/3 - 0] = π/12

Answer 3

See the video.

https://www.integral-calculator.com/ is an online integral calculator with steps - so you can check your answers.