Question

Integrate (sinx)^4(cosx)^2

Answer 1

wolfram aplha i got pretty far n not enough space to see where i did the mistake.

Answer 2

its REALLY UGLY you set into cos^3(2x) and have to combine terms

i got to this point
3/8+(-7/8cos(2x))+(5/8cos(2x))+(-1/8cos^3(2x))

Answer 3

3/8+(-7/8cos(2x))+(5/8cos^2(2x))+(-1/8cos^3(2x))

Answer 4

-1/8cos^3(2x)=-1/8(cos(2x)(1-sin^2(2x))
-1/8cos(2x)+1/8cos(2x)sin^2(2x)

Answer 5

cos^2(2x)=(1/2+1/2cos(4x))

Answer 6

3/8+(-7/8cos(2x))+(5/8(1/2+1/2cos(4x)))+(-1/8cos(2x)+1/8cos(2x)sin^2(2x))

Answer 7

3/8+5/16=11/16
-cos(2x)
+1/2cos(4x)
1/8cos(2x)sin^2(2x)\[1/8\int\limits_{}^{}\cos^2(2x)\sin^2(2x)dx\]+\[\int\limits_{}^{}1/2\cos(4x)dx\]+\[-\int\limits_{}^{}\cos(2x)dx\]+\[\int\limits_{}^{}11/16dx\]


\[\int\limits_{}^{}1/8\cos(2x)\sin^2(2x)=1/48(\sin^3(2x))\]
\[\int\limits_{}^{}1/2\cos(4x)=1/8\sin(4x)\]
\[\int\limits_{}^{}-\cos(2x)=-1/2\sin(2x)\]
\[\int\limits_{}^{}11/16=11/16+C\]


1/48sin^3(2x)+1/8sin(4x)-1/2sin(2x)+11/16x+C

Answer 8

which is not whats on wolfram ha anyways