Example1: Find ∫sin⁴x dx

∫sin⁴x dx =∫(sin²x)² dx

                     1

                =∫— (1–cos2x)² dx

                     4

                     1

                =∫— [1– 2cos2x + cos²(2x) ] dx

                     4

                      1                          1

                = ∫— [1– 2cos2x + —(1+ cos4x)] dx

                      4                          2

                         1      1                 1      1

                = ∫[ — – — cos2x + — + — cos4x)] dx

                         4      2                 8      8

                     1            1                       1             1

                =  —∫dx – —∫cos2x .dx + — ∫dx + — ∫ cos4x .dx

                     4            2                       8             8

                     1         1                1          1

                =  — x – — sin2x + — x + — sin4x + c

                     4         4                8         32

                     3         1                 1

                =  — x – — sin2x +  — sin4x + c

                     8         4                32