What is the integral of sin^4(5x)*cos^2(5x) ?

The integral of sin^4(5x)*cos^2(5x) is 1/480 (30x-3sin(10x)-4sin^3(5x)cos(5x)+16sin^5(5x)cos(5x))+C

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integral of sin^4(5x)*cos^2(5x)

\int sin^{4} (5 x) \cdot cos^{2} (5 x) d x

\frac{1}{480} (30 x - 3 sin (10 x) - 4 sin^{3} (5 x) cos (5 x) + 16 sin^{5} (5 x) cos (5 x)) + C

Solution steps

\int sin^{4} (5 x) cos^{2} (5 x) d x

Apply u-substitution

: \int sin^{4} (u) cos^{2} (u) \frac{1}{5} d u

= \int sin^{4} (u) cos^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sin^{4} (u) cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int sin^{4} (u) (1 - sin^{2} (u)) d u

Expand

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

= \frac{1}{5} \cdot \int sin^{4} (u) - sin^{6} (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}{5} (\int sin^{4} (u) d u - \int sin^{6} (u) d u)

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

\int sin^{6} (u) d u = - \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))

= \frac{1}{5} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)) - (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u)))))

Substitute back

u = 5 x

= \frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x)) - (- \frac{cos ( 5 x ) sin ^{5} ( 5 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x)))))

Simplify

\frac{1}{5} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x)) - (- \frac{cos ( 5 x ) sin ^{5} ( 5 x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (5 x) cos (5 x) + \frac{3}{8} (5 x - \frac{1}{2} sin (2 \cdot 5 x))))) : \frac{1}{480} (30 x - 3 sin (10 x) - 4 sin^{3} (5 x) cos (5 x) + 16 sin^{5} (5 x) cos (5 x))

= \frac{1}{480} (30 x - 3 sin (10 x) - 4 sin^{3} (5 x) cos (5 x) + 16 sin^{5} (5 x) cos (5 x))

Add a constant to the solution

= \frac{1}{480} (30 x - 3 sin (10 x) - 4 sin^{3} (5 x) cos (5 x) + 16 sin^{5} (5 x) cos (5 x)) + C

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What is the integral of sin^4(5x)*cos^2(5x) ?
The integral of sin^4(5x)*cos^2(5x) is 1/480 (30x-3sin(10x)-4sin^3(5x)cos(5x)+16sin^5(5x)cos(5x))+C