Anti derivative for : cos^2xsin^3x

Question

Find the antiderivative for cosine sqaured x multiplied by sin to the cubed x

(cosx)^2 times (sinx)^3

Answer 1

   ∫ cos²x·sin³x  dx = ∫ cos²x·sinx·sin²x  dx

Recall:     sin²x + cos²x = 1     ==>     sin²x = 1 - cos²x

   ∫ cos²x·sinx·(1 - cos²x)  dx  =  ∫ (cos²x·sinx - cos⁴x·sinx)  dx

   =  ∫ cos²x·sinx  dx  -  ∫ cos⁴x·sinx  dx

Using u-substitution, let:     u = cosx   , thus,   du = -sinx  dx

 ==>   ∫ cos²x·sinx  dx  =  ∫ -1·cos²x·(-1·sinx  dx)  =  ∫ -(u)²·(du)  =  ∫ -u²  du  

             =  -(u)³/3  =  -(cosx)³/3  =  -cos³x/3 

 ==>   ∫ cos⁴x·sinx  dx  =  ∫ -1·cos⁴x·(-1·sinx  dx)  =  ∫ -(u)⁴·(du)  =  ∫ -u⁴  du

             =  -(u)⁵/5  =  -(cosx)⁵/5  =  -cos⁵x/5

∫ cos²x·sinx  dx  -  ∫ cos⁴x·sinx  dx  =  (-cos³x/3) - (-cos⁵x/5)

                                                    =  -cos³x/3  +  cos⁵x/5  =  cos⁵x/5 - cos³x/3

Answer 2

Maro.

This integral requires a trig substitution and a u-substitution.  Its the trig substitution that is difficult.  When the sine is to an odd power, you want to rewrite the sin term as an even power of sin x times sin x, here, (sin^2 x) (sin x).  Then substitute 1-cos^2 x for sin^2 of x and then distribute.  In this case you now need to integrate this: [cos^2 x][1-cos^2 x][sin x].  Distribute this out and you will get [cos^2 x][sin x] - [cos^4 x][sin x], with this you can use sum-rule, u-substitution and the power rule to solve each term, which is fairly straightforward.  Feel free to email or ask a follow-up question if this does not make sense.  John