[tex]integral \: dari \: {cos}^{5} x \: {sin}^{4} x \: = ...[/tex]
Matematika
laurays2
Pertanyaan
[tex]integral \: dari \: {cos}^{5} x \: {sin}^{4} x \: = ...[/tex]
1 Jawaban
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1. Jawaban Anonymoux
[tex] \int\limits cos^5x\ sin^4x \, dx =\\ \int\limits cosx.cos^4x.sin^4x\ dx\\ \int\limits sin^4x.cos^4x.cosx\ dx\\ \int\limits sin^4x.(1-sin^2x).(1-sin^2x).cosx\ dx\\ \int\limits sin^4x.(1-sin^2x).(1-sin^2x).d(sinx)\\ \int\limits u^4(1-u^2).(1-u^2)du\\\int\limits\ u^4(1-2u^2+u^4)du\\ \int\limits (u^4-2u^6+u^8)du\\ \frac {u^5}{5} -\frac {2u^7}{7} +\frac {u^9}{9} + C\\ \\ \frac {sin^5x}{5} -\frac {2sin^7x}{7}+\frac {sin^9x}{9}\ +C\\ \\ CATATAN:\\ sin^2x=1-cos^2x\\ cos^2x=1-sin^2x\\ cos^4x=(1-sin^2x)^2\\ [/tex]
[tex]cos^4x=(1-sin^2x).(1-sin^2x)\\ \\ dsinx=cosx.dx\\ [/tex]
jika template sudah sama maka bisa diintegral,itu sebabnya cos x diubah ke sin x