Find ∫sin4xdx.
Find the integration of the given function.
∫sin4xdx=∫(sin2x)2dx
=∫1−cos2x22dx...[sin2x=1-cos2x2]=14∫(1−cos2x)2dx=14∫(1−2cos2x+cos22x)dx...[(a-b)2=a2-2ab+b2]=14∫1–2cos2x+1+cos4x2dx=14∫dx–24∫cos2xdx+18∫(1+cos4x)dx=x4–12sin2x2+x8+18sin4x4+C
⇒∫sin4xdx=38x+132sin(4x)−14sin(2x)+C
Hence, the required answer is ∫sin4xdx=38x+132sin(4x)−14sin(2x)+C
(i) Find the integral: ∫dxx2−6x+13 (ii) Find the integral: ∫dx3x2+13x−10 (iii) Find the integral: ∫dx√5x2−2x