What is the integral of sin^4(7x-2) ?

The integral of sin^4(7x-2) is 1/7 (-1/4 sin^3(7x-2)cos(7x-2)+3/8 (7x-2-1/2 sin(2(7x-2))))+C

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integral of sin^4(7x-2)

\int sin^{4} (7 x - 2) d x

\frac{1}{7} (- \frac{1}{4} sin^{3} (7 x - 2) cos (7 x - 2) + \frac{3}{8} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2)))) + C

Solution steps

\int sin^{4} (7 x - 2) d x

Apply u-substitution

= \int sin^{4} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{7} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= \frac{1}{7} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 7 x - 2

= \frac{1}{7} (- \frac{cos ( 7 x - 2 ) sin ^{3} ( 7 x - 2 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2))))

Simplify

\frac{1}{7} (- \frac{cos ( 7 x - 2 ) sin ^{3} ( 7 x - 2 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2)))) : \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x - 2) cos (7 x - 2) + \frac{3}{8} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2))))

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x - 2) cos (7 x - 2) + \frac{3}{8} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2))))

Add a constant to the solution

= \frac{1}{7} (- \frac{1}{4} sin^{3} (7 x - 2) cos (7 x - 2) + \frac{3}{8} (7 x - 2 - \frac{1}{2} sin (2 (7 x - 2)))) + C

3 y^{^{''}} - 2 y^{^{'}} - 8 y = 0

3 y^{^{''}} + 7 y^{^{'}} + 2 y = 3 x^{2} + 3 x + 1, y (0) = 0, y^{^{'}} (0) = 0

7 x sin (x)

\int \frac{8}{x ^{2} - 2 x + 2} d x

\frac{\partial}{\partial x} (x y - x - 3 y)

What is the integral of sin^4(7x-2) ?
The integral of sin^4(7x-2) is 1/7 (-1/4 sin^3(7x-2)cos(7x-2)+3/8 (7x-2-1/2 sin(2(7x-2))))+C