The question is ∫3cos25x dx. I tried to solve it 2 ways and got two different answers.

Method A.
∫3cos25x dx =3∫cos25x dx

u=5x
3∫cos2u (1/5)du/dx dx
The 2 dx cancel out leaving
3∫cos2u (1/5)du=
3/5 ∫cos2u 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
∫3cos25x dx =3∫cos25x 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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eumyang
Homework Helper
The question is ∫3cos25x dx. I tried to solve it 2 ways and got two different answers.

Method A.
∫3cos25x dx =3∫cos25x dx

u=5x
3∫cos2u (1/5)du/dx dx
The 2 dx cancel out leaving
3∫cos2u (1/5)du=
3/5 ∫cos2u du=
3/5 ∫(1/2 + 1/2 cos 2u) du=
3/5( (1/2x)+∫1/2 cos 2u du)
Here, it should be $\frac{1}{2}u$. Then substitute the 5x in for u.

Ray Vickson