NettetIntegrate the function xsin −1x Hard Solution Verified by Toppr Let I=∫xsin −1xdx Taking sin −1x as first function and x as second function and integrating by parts, we obtain I=sin −1x∫xdx−∫{(dxd sin −1x)∫xdx}dx =sin −1x( 2x 2)−∫ 1−x 21 ⋅ 2x 2dx = 2x 2sin −1x+ 21∫ 1−x 2−x 2 dx = 2x 2sin −1x+ 21∫{ 1−x 21−x 2− 1−x 21 }dx Nettet15. okt. 2015 · One way is to just integrate by parts, ∫ x ( 1 + x 2) 3 / 2 arctan x = − 1 1 + x 2 arctan x + ∫ 1 ( 1 + x 2) 3 / 2 d x = − 1 1 + x 2 arctan x + x 1 + x 2. Here, the last step might seem mystery. I know it by heart, but one way to conclude it is: 1 ( 1 + x 2) 3 / 2 = ( 1 + x 2) − x 2 ( 1 + x 2) 3 / 2 = 1 1 + x 2 − x 2 ( 1 + x 2) 3 / ...

∫xtan-1x.dx - Mathematics and Statistics - Shaalaa.com

NettetLearn how to solve integral calculus problems step by step online. Find the integral of x^21/2x. Find the integral. When multiplying exponents with same base you can add the exponents: \frac{1}{2}x^2x. The integral of a function times a constant (\frac{1}{2}) is equal to the constant times the integral of the function. Apply the power rule for integration, … Nettet16. mar. 2024 · Misc 23 - Chapter 7 Class 12 Integrals (Term 2) Last updated at March 16, 2024 by Teachoo. Get live Maths 1-on-1 Classs - Class 6 to 12. Book 30 minute class for ₹ 499 ₹ 299. Transcript. force field nexus hexxit 2

Evaluate: ∫ x(tan^-1x)^2 dx, x ∈[0,1] - Sarthaks

Nettet26. aug. 2016 · Use tan2x = sec2x −1 first. Explanation: xtan2x = xsec2x −x −∫xdx = − x2 2 ∫xsec2xdx Let u = x and dv = sec2xdx, so that du = dx and v = tanx to get ∫xsec2xdx = xtanx − ∫tanxdx. Now integrate tanx = sinx cosx using substitution u = cosx. ∫xsec2xdx = xtanx − ( − ln cosx ) = xtanx + ln cosx Finish by putting it all together. Nettet12. nov. 2024 · Best answer Let x = tanθ ⇒ dx = sec2θ dθ Then, I = ∫ tanθ tan−1(tanθ) (1+tan2θ)3/2 ∫ t a n θ t a n − 1 ( t a n θ) ( 1 + t a n 2 θ) 3 / 2 sec2θ dθ = ∫ θ tanθ sec3θ ∫ θ t a n θ s e c 3 θ sec2θ dθ = ∫ θ tanθ secθ ∫ θ t a n θ s e c θ dθ = ∫ θ sinθ dθ ∫ θ s i n θ d θ Nettet20. feb. 2024 · Explanation: We want to solve I = ∫tan−1(x)dx Use integration by parts / partial integration ∫udv = uv − ∫vdu Let u = tan−1(x) and dv = 1dx Then du = 1 x2 + 1 dx and v = x I = tan−1(x)x − ∫ x x2 +1 dx Make a substitution u = x2 +1 ⇒ du dx = 2x I = tan−1(x)x − 1 2 ∫ 1 u du = tan−1(x)x − 1 2ln(u) +C Substitute back u = x2 + 1 force field model of change management