What is the integral of cos^6(x)sin(2x) ?

The integral of cos^6(x)sin(2x) is -1/4 cos^8(x)+C

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integral of cos^6(x)sin(2x)

\int cos^{6} (x) sin (2 x) d x

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

Solution steps

\int cos^{6} (x) sin (2 x) d x

Rewrite using trig identities

= \int 2 cos^{7} (x) sin (x) d x

Take the constant out:

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

= 2 \cdot \int cos^{7} (x) sin (x) d x

Apply u-substitution

: \int - u^{7} d u

= 2 \cdot \int - u^{7} d u

Take the constant out:

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

= 2 (- \int u^{7} d u)

Apply the Power Rule

: \frac{u ^{8}}{8}

= 2 (- \frac{u ^{8}}{8})

Substitute back

u = cos (x)

= 2 (- \frac{cos ^{8} ( x )}{8})

Simplify

2 (- \frac{cos ^{8} ( x )}{8}) : - \frac{1}{4} cos^{8} (x)

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

Add a constant to the solution

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

(2 y + 2 x - 7) d x + (4 y + 2 x - 1) d y = 0

\int (\frac{1}{x ^{2} - 1}) d x

ln (1 + x), 5

2 x^{2} - 4, x^{2}

What is the integral of cos^6(x)sin(2x) ?
The integral of cos^6(x)sin(2x) is -1/4 cos^8(x)+C