What is #int sin^3x+3sin^2x+2sinx-5 dx#?
2 Answers
Explanation:
Using the linearity of the integral:
Solve the integrals separately:
Putting the partial solutions together:
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Explanation:
The question becomes
# =>I= I_1+I_2+I_3+I_4+C#
where#C# is a constant of integration.
Now
Let us substitute
#=>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#