Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 11

To solve the indefinite integral, we follow int f(x) dx = F(x) +C
where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
For the given integral problem: int x sin^2(x) dx, we may apply integration by parts: int u *dv = uv - int v *du .
We may let:
u = x then du =1 dx or dx
dv= sin^2(x) dx then v = x/2 - sin(2x)/4
Note: From the table of integrals, we have int sin^2(ax) dx = x/2 - sin(2ax)/(4a) . We apply this on v =int dv =intsin^2(x) dx where a =1 .
Applying the formula for integration by parts, we have:
int x sin^2(x) dx= x*(x/2 - sin(2x)/4 ) - int (x/2 - sin(2x)/4 ) dx
=x^2/2 - (xsin(2x))/4 - int (x/2 - sin(2x)/4 ) dx
For the integral: int (x/2 - sin(2x)/4 ) dx , we may apply the basic integration property: : int (u-v) dx = int (u) dx - int (v) dx .

int (x/2 - sin(2x)/4 ) dx =int (x/2) dx -int sin(2x)/4 ) dx
= 1/2 int x dx - 1/4 int sin(2x) dx .

Apply the Power rule for integration:
int x^n dx = x^(n+1)/(n+1) +c
1/2 int x dx = 1/2*x^(1+1)/(1+1)
= 1/2* x^2/2
= x^2/4
Apply the basic integration formula for sine function: int sin(u) du = -cos(u) +C .
Let: u =2x then du = 2 dx or (du)/2 = dx .
1/4 int sin(2x) dx = 1/4 int sin(u) * (du)/2
= 1/4 *1/2 int sin(u) du
= 1/8 (-cos(u))
= -cos(u)/8
Plug-in u = 2x on -cos(u)/8 , we get: 1/4 int sin(2x) dx =-cos(2x)/8 .
Combining the results, we get:
int (x/2 - sin(2x)/4 ) dx =x^2/4 - (-cos(2x)/8) +C
=x^2/4+ cos(2x)/8 +C
Then, the complete indefinite integral will be:
int x sin^2(x) dx=x^2/2 - (xsin(2x))/4 - int (x/2 - sin(2x)/4 ) dx
=x^2/2 - (xsin(2x))/4 -(x^2/4+ cos(2x)/8) +C
=x^2/2 - (xsin(2x))/4 - x^2/4 - cos(2x)/8 +C
= (x^2)/4- (xsin(2x))/4- cos(2x)/8 +C