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The question is ∫3cos

Method A.

∫3cos

u=5x

3∫cos

The 2 dx cancel out leaving

3∫cos

3/5 ∫cos

3/5 ∫(1/2 + 1/2 cos 2u) du=

3/5( (1/2x)+∫1/2 cos 2u du)=

3/5( (1/2x)+1/2∫ cos 2u du)=

3/5( (1/2x)+1/2(1/2sin2u))=

3/5( (1/2x)+1/4sin2u)=

3/10x + 3/20 sin 2u=

3/10x + 3/20 sin 10x +C

Method B

∫3cos

3∫(1/2 + 1/2cos 2(5x)) dx=

3∫(1/2 + 1/2cos 10x) dx=

3∫(1/2(1+cos10x)) dx=

3/2 ∫(1+cos10x) dx=

3/2 (∫1 dx+ ∫cos10x dx)=

3/2(x+∫cos10x dx)=

3/2x+(3/2)(1/10 sin 10x)=

3/2x+ 3/20 sin10x +C

Did I do something wrong in either of the methods? thanks!

^{2}5x dx. I tried to solve it 2 ways and got two different answers.Method A.

∫3cos

^{2}5x dx =3∫cos^{2}5x dxu=5x

3∫cos

^{2}u (1/5)du/dx dxThe 2 dx cancel out leaving

3∫cos

^{2}u (1/5)du=3/5 ∫cos

^{2}u du=3/5 ∫(1/2 + 1/2 cos 2u) du=

3/5( (1/2x)+∫1/2 cos 2u du)=

3/5( (1/2x)+1/2∫ cos 2u du)=

3/5( (1/2x)+1/2(1/2sin2u))=

3/5( (1/2x)+1/4sin2u)=

3/10x + 3/20 sin 2u=

3/10x + 3/20 sin 10x +C

Method B

∫3cos

^{2}5x dx =3∫cos^{2}5x dx=3∫(1/2 + 1/2cos 2(5x)) dx=

3∫(1/2 + 1/2cos 10x) dx=

3∫(1/2(1+cos10x)) dx=

3/2 ∫(1+cos10x) dx=

3/2 (∫1 dx+ ∫cos10x dx)=

3/2(x+∫cos10x dx)=

3/2x+(3/2)(1/10 sin 10x)=

3/2x+ 3/20 sin10x +C

Did I do something wrong in either of the methods? thanks!

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