What is int sin^3x+3sin^2x+2sinx-5 dx∫sin3x+3sin2x+2sinx−5dx?

2 Answers
Andrea S.
Apr 11, 2018

int (sin^3x +3sin^2x +2sinx -5) dx = -(cosx(2sin^2x +9sinx +16) + 21x)/6 +C∫(sin3x+3sin2x+2sinx−5)dx=−cosx(2sin2x+9sinx+16)+21x6+C

Explanation:

Using the linearity of the integral:

int (sin^3x +3sin^2x +2sinx -5) dx = int sin^3xdx +3int sin^2xdx+2 int sinxdx -5 int dx∫(sin3x+3sin2x+2sinx−5)dx=∫sin3xdx+3∫sin2xdx+2∫sinxdx−5∫dx

Solve the integrals separately:

(1)(1)

int dx = x +c_1∫dx=x+c1

(2)(2)

int sinxdx = -cosx + c_2∫sinxdx=−cosx+c2

(3)(3)

int sin^2x = int (1-cos2x)/2 dx ∫sin2x=∫1−cos2x2dx

int sin^2x = 1/2 int dx -1/2 int cos2x dx ∫sin2x=12∫dx−12∫cos2xdx

int sin^2x = x/2 -(sin2x)/4 +c_3 ∫sin2x=x2−sin2x4+c3

int sin^2x = x/2 -(sinxcosx)/2 +c_3 ∫sin2x=x2−sinxcosx2+c3

(4)(4)

int sin^3xdx = int (1-cos^2x) sinxdx∫sin3xdx=∫(1−cos2x)sinxdx

int sin^3xdx = int sinx dx + int cos^2x d(cosx)∫sin3xdx=∫sinxdx+∫cos2xd(cosx)

int sin^3xdx = -cosx + (cos^3x)/3 +c_4∫sin3xdx=−cosx+cos3x3+c4

Putting the partial solutions together:

int (sin^3x +3sin^2x +2sinx -5) dx = -cosx + (cos^3x)/3 + (3x)/2 -(3sinxcosx)/2 -2cosx -5x +C∫(sin3x+3sin2x+2sinx−5)dx=−cosx+cos3x3+3x2−3sinxcosx2−2cosx−5x+C

int (sin^3x +3sin^2x +2sinx -5) dx = (cos^3x)/3 -(3sinxcosx)/2 -3cosx - (7x)/2 +C∫(sin3x+3sin2x+2sinx−5)dx=cos3x3−3sinxcosx2−3cosx−7x2+C

int (sin^3x +3sin^2x +2sinx -5) dx = (2cos^3x -9sinxcosx -18cosx - 21x)/6 +C∫(sin3x+3sin2x+2sinx−5)dx=2cos3x−9sinxcosx−18cosx−21x6+C

int (sin^3x +3sin^2x +2sinx -5) dx = (cosx(2cos^2x -9sinx -18) - 21x)/6 +C∫(sin3x+3sin2x+2sinx−5)dx=cosx(2cos2x−9sinx−18)−21x6+C

int (sin^3x +3sin^2x +2sinx -5) dx = (cosx(2- 2sin^2x -9sinx -18) - 21x)/6 +C∫(sin3x+3sin2x+2sinx−5)dx=cosx(2−2sin2x−9sinx−18)−21x6+C

int (sin^3x +3sin^2x +2sinx -5) dx = -(cosx(2sin^2x +9sinx +16) + 21x)/6 +C∫(sin3x+3sin2x+2sinx−5)dx=−cosx(2sin2x+9sinx+16)+21x6+C

A08
Apr 11, 2018

There is an error while posting the question. All terms must be multiplied with dxdx, not only the last as given.

Explanation:

The question becomes

int\ (sin^3x +3sin^2x +2sinx -5) dx
=>I= int\ sin^3x\ dx +3int\ sin^2x\ dx+2 int\ sinx\ dx -5 int\ dx

=>I= I_1+I_2+I_3+I_4+C
where C is a constant of integration.

Now I_1=int\ sin^3x\ dx
=>I_1= int\ (1-cos^2x) sinx\ dx

Let us substitute u=cosx->du=-sinx\ dx

=>I_1= int\ (u^2-1)\ du
=>I_1= u^3/3-u

Reversing the substitution we get

I_1 = (cos^3x)/3 -cosx

I_2=3int\ sin^2x\ dx
=>I_2 =3 int\ (1-cos2x)/2\ dx
=>I_2 =3[ 1/2 int\ dx -1/2 int\ cos2x dx]
=>I_2= 3[x/2 -(sin2x)/4 ]
=>I_2= (3x)/2 -3/4(sin2x)

I_3=2int\ sinx\ dx = -2cosx

I_4=-5int\ dx = -5x

Therefore

I=( (cos^3x)/3 -cosx)+((3x)/2 -3/4(sin2x))+( -2cosx)+(-5x)+C
I=-(9sin2x-4cos^3x+36cosx+42x)/12+C_1

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