What is the integral of xsin^2(2x^2)cos^2(2x^2) ?

The integral of xsin^2(2x^2)cos^2(2x^2) is 1/16 (x^2-1/8 sin(8x^2))+C

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integral of xsin^2(2x^2)cos^2(2x^2)

\int x sin^{2} (2 x^{2}) cos^{2} (2 x^{2}) d x

\frac{1}{16} (x^{2} - \frac{1}{8} sin (8 x^{2})) + C

Solution steps

\int x sin^{2} (2 x^{2}) cos^{2} (2 x^{2}) d x

Apply u-substitution

: \int \frac{sin ^{2} ( 2 u ) cos ^{2} ( 2 u )}{2} d u

= \int \frac{sin ^{2} ( 2 u ) cos ^{2} ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin^{2} (2 u) cos^{2} (2 u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{1}{8} (1 - cos (8 u)) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int 1 - cos (8 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{8} (\int 1 d u - \int cos (8 u) d u)

\int 1 d u = u

\int cos (8 u) d u = \frac{1}{8} sin (8 u)

= \frac{1}{2} \cdot \frac{1}{8} (u - \frac{1}{8} sin (8 u))

Substitute back

u = x^{2}

= \frac{1}{2} \cdot \frac{1}{8} (x^{2} - \frac{1}{8} sin (8 x^{2}))

Simplify

\frac{1}{2} \cdot \frac{1}{8} (x^{2} - \frac{1}{8} sin (8 x^{2})) : \frac{1}{16} (x^{2} - \frac{1}{8} sin (8 x^{2}))

= \frac{1}{16} (x^{2} - \frac{1}{8} sin (8 x^{2}))

Add a constant to the solution

= \frac{1}{16} (x^{2} - \frac{1}{8} sin (8 x^{2})) + C

\frac{x ^{2} + 1}{x ^{2} - 1}

y = x^{2} + 5, (4, 21)

u \to 0 lim (ln (\frac{u + 1}{u}))

y = - x^{3} + 2 x^{2} - 4, at x = - 1

(- 4.3) (5.3)

What is the integral of xsin^2(2x^2)cos^2(2x^2) ?
The integral of xsin^2(2x^2)cos^2(2x^2) is 1/16 (x^2-1/8 sin(8x^2))+C