Question

integration of root of x square + a square

Answer 1

Let x=sinθ⟹dx=cosθ dθx=sin⁡θ⟹dx=cos⁡θ dθ

∴∫1−x2√x dx=∫1−sin2θ√sinθcosθ dθ∴∫1−x2x dx=∫1−sin2⁡θsin⁡θcos⁡θ dθ

=∫cosθsinθcosθ dθ=∫cos⁡θsin⁡θcos⁡θ dθ

=∫cos2θsinθ dθ=∫cos2⁡θsin⁡θ dθ

=∫1−sin2θsinθ dθ=∫1−sin2⁡θsin⁡θ dθ

=∫cscθ dθ−∫sinθ dθ=∫csc⁡θ dθ−∫sin⁡θ dθ

=ln|cscθ−cotθ|+cosθ+C=ln⁡|csc⁡θ−cot⁡θ|+cos⁡θ+C

=ln∣∣∣1x−1−x2√x∣∣∣+1−x2−−−−−√+

Step-by-step explanation:

I HOPE IT'S HELPFUL for you