`int xsin^2x dx` Find the indefinite integral

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...

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`