What is the integral of xcos^4(x^2) ?

The integral of xcos^4(x^2) is 1/2 (cos^3(x^2)sin(x^2)+3/8 (x^2-1/4 sin(4x^2)))+C

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integral of xcos^4(x^2)

\int x cos^{4} (x^{2}) d x

\frac{1}{2} (cos^{3} (x^{2}) sin (x^{2}) + \frac{3}{8} (x^{2} - \frac{1}{4} sin (4 x^{2}))) + C

Solution steps

\int x cos^{4} (x^{2}) d x

Apply u-substitution

: \int \frac{cos ^{4} ( u )}{2} d u

= \int \frac{cos ^{4} ( 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 cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{2} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

: cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u

= \frac{1}{2} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{2} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = x^{2}

= \frac{1}{2} (cos^{3} (x^{2}) sin (x^{2}) - (- \frac{3}{8} (x^{2} - \frac{1}{4} sin (4 x^{2}))))

Simplify

= \frac{1}{2} (cos^{3} (x^{2}) sin (x^{2}) + \frac{3}{8} (x^{2} - \frac{1}{4} sin (4 x^{2})))

Add a constant to the solution

= \frac{1}{2} (cos^{3} (x^{2}) sin (x^{2}) + \frac{3}{8} (x^{2} - \frac{1}{4} sin (4 x^{2}))) + C

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 2 t}

x^{'} (t) = \frac{1}{3 x ^{2} + 1}, x (0) = 1

\frac{\partial}{\partial x} (e^{- 6 t} cos (π x))

\frac{\partial}{\partial y} (9 z e^{x yz})

\int_{0}^{1} (e^{- 0.1 t}) d t

What is the integral of xcos^4(x^2) ?
The integral of xcos^4(x^2) is 1/2 (cos^3(x^2)sin(x^2)+3/8 (x^2-1/4 sin(4x^2)))+C