Vikas TU
Last Activity: 8 Years ago
Apply By part integeration ---->
Let the first function be unity and second one root sinx.
integration of square root of sin x dx
integration (sin x)^1/2 dx = [{(sin x)^3/2) / (3/2)}* cos x
= [2cos x.sin x.(sin x)^1/2] / 3
= [sin2x.(sin x)^1/2] / 3
AAp;;y the limits
2nd method:
let sinx = t
cos x dx= dt
dx=dt/cosx
= dt / (1-t^2)^1/2
Required: integrate(sinx ^ 1/2 dx)
integrate( [t^1/2 ]/[(1-t^2)^1/2 dt )
sorry ,I cant solve it by this method,
BUT , I can give u the answer directly.
ANSWER: 2/3[ ( sinx)^3/2]/cosx
2/3 * (sinx)^3/2 * secx