How to integrate ∫ sin 2 ⁡ θ cos 5 ⁡ θ d θ

Question

How to integrate   ∫                              sin                      2                          ⁡        θ                              cos                      5                          ⁡        θ              d    θ

namotanimfc · Accepted Answer

∫                              sin          2                ⁡        θ                              cos          5                ⁡        θ                  d    θ  =  ∫            1                        cos          5                ⁡        θ                  d    θ  −  ∫            1                        cos          3                ⁡        θ                  d    θLet       I    n    :=  ∫      cos    n    ⁡  θ      d    θwe have                              I          n                =        ∫                  cos                      n            −            1                          ⁡        θ                  d                sin        ⁡        θ                            =                  cos                      n            −            1                          ⁡        θ        sin        ⁡        θ        −        ∫        sin        ⁡        θ                  d                          cos                      n            −            1                          ⁡        θ                                          =                  cos                      n            −            1                          ⁡        θ        sin        ⁡        θ        +        (        n        −        1        )        ∫                  cos                      n            −            2                          ⁡        θ                  sin          2                ⁡        θ                  d                θ                                          =                  cos                      n            −            1                          ⁡        θ        sin        ⁡        θ        +        (        n        −        1        )        (                  I                      n            −            2                          −                  I          n                )            hence,   n      I    n    =  (  n  −  1  )      I          n      −      2        +      cos          n      −      1        ⁡  θ  sin  ⁡  θAlso, we have       I          −      1        =  ∫  sec  ⁡  θ      d    θ  =  ln  ⁡      |    sec  ⁡  θ  +  tan  ⁡  θ      |    +  CSo,       I          −      3        =                    ln        ⁡                  |                sec        ⁡        θ        +        tan        ⁡        θ                  |                +                  sec          2                ⁡        θ        sin        ⁡        θ            2        +  C and       I          −      5        =                    3                  I                      −            3                          +                  sec          4                ⁡        θ        sin        ⁡        θ            4        ∴  ∫                              sin          2                ⁡        θ                              cos          5                ⁡        θ                  d    θ  =      I          −      5        −      I          −      3        =                    −        ln        ⁡                  |                sec        ⁡        θ        +        tan        ⁡        θ                  |                −                  sec          2                ⁡        θ        sin        ⁡        θ        +        2                  sec          4                ⁡        θ        sin        ⁡        θ            8        +  C

How to integrate int(sin^2theta)/(cos^5theta)

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