Question

What is the answer of `intcos^3(sqrtx)dx`?

Answer 1

`intcos^3(sqrtx)dx = 3/2(sqrtxsin(sqrtx)+cos(sqrtx))+ 1/18(3usin(3sqrtx)+ cos(3sqrtx))+ C`

Explanation:

Let `u = sqrtx`, the `du = 1/(2sqrtx)dx` or `dx = 2sqrtxdu = 2u du` then:

`intcos^3(sqrtx)dx = 2intucos^3(u)du`

Separate into `cos(u)cos^2(u)`:

`intcos^3(sqrtx)dx = 2intucos(u)cos^2(u)du`

Use the identity `cos^2(u) = 1/2(cos(2u)+1)`:

`intcos^3(sqrtx)dx = intucos(u)(cos(2u)+1)du`

Use the distributive property:

`intcos^3(sqrtx)dx = intucos(u)cos(2u)+ucos(u)du`

Use the identity `cos(A)cos(B) = 1/2(cos(A-B)+cos(A+B)` where `A = 2u and B = u`:

`intcos^3(sqrtx)dx = int 1/2u(cos(u)+ cos(3u))+ucos(u)du`

Combine like terms:

`intcos^3(sqrtx)dx = int 3/2ucos(u)+ 1/2ucos(3u)du`

Separate into two integrals:

`intcos^3(sqrtx)dx = 3/2int ucos(u)du+ 1/2int ucos(3u)du`

Both integrals can be integrated by parts:

`intcos^3(sqrtx)dx = 3/2(usin(u)+cos(u))+ 1/18(3usin(3u)+ cos(3u))+ C`

Reverse the substitution `u = sqrtx`:

`intcos^3(sqrtx)dx = 3/2(sqrtxsin(sqrtx)+cos(sqrtx))+ 1/18(3usin(3sqrtx)+ cos(3sqrtx))+ C`

Answer 2

`=sqrtx/6(sin3sqrtx+9sinsqrtx)+1/18(cos3sqrtx+27cossqrtx)+C`
Hint:
`cos3x=4cos^3x-3cosx`

`=>4cos^3x=cos3x+3cosx`

`=>cos^3x=1/4(cos3x+3cosx)...to(A)`

Explanation:

We know that,

`(1)intcoskx=1/ksinkx+c`

`(2)intsinkx=1/k(-coskx)+c`

Here, `I=intcos^3(sqrtx)dx`

Let , `sqrtx=t=>x=t^2=>dx=2tdt`

`:.I=intcos^3t*2tdt`

`:.I=2inttcos^3tdt`

Using `(A)` ,we get

`I=2xx1/4int t(cos3t+3cost)dt`

`=1/2inttcos3tdt+3/2inttcostdt`

`"Using "color(blue)"Integration by Parts"`, in both integrals,

`I=1/2[tintcos3tdt-int(d/(dt)(t)intcos3tdt)dt]`

`color(white)(......)+3/2[tcostdt-int(d/(dt)(t)intcostdt)dt]toApply(1)`

`I=1/2[t((sin3t)/3)-int(1)(sin3t)/3dt]`

`color(white)(.....)+3/2[t(sint)-int(1)sintdt]`

`I=1/6[tsin3t-intsin3tdt]+3/2[tsint-intsintdt]`

`=1/6[tsin3t-((-cos3t)/3)]+3/2[tsint-(-cost)]+C`

`=t/6sin3t+1/18cos3t+(3t)/2sint+3/2cost+C,`

`=t/6sin3t+(3t)/2sint+1/18cos3t+3/2cost+C`

`=t/6(sin3t+9sint)+1/18(cos3t+27cost)+C`

Substituting back, `t=sqrtx`

`=sqrtx/6(sin3sqrtx+9sinsqrtx)+1/18(cos3sqrtx+27cossqrtx)+C`