What is the integral of cos^4(7x+8) ?

The integral of cos^4(7x+8) is 1/224 (32cos^3(7x+8)sin(7x+8)+3(28x+32-sin(4(7x+8))))+C

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integral of cos^4(7x+8)

\int cos^{4} (7 x + 8) d x

\frac{1}{224} (32 cos^{3} (7 x + 8) sin (7 x + 8) + 3 (28 x + 32 - sin (4 (7 x + 8)))) + C

Solution steps

\int cos^{4} (7 x + 8) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{7} d u

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

Simplify

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

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

Substitute back

u = 7 x + 8

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

Simplify

\frac{1}{7} (cos^{3} (7 x + 8) sin (7 x + 8) - (- \frac{3}{8} (7 x + 8 - \frac{1}{4} sin (4 (7 x + 8))))) : \frac{1}{224} (32 cos^{3} (7 x + 8) sin (7 x + 8) + 3 (28 x + 32 - sin (4 (7 x + 8))))

= \frac{1}{224} (32 cos^{3} (7 x + 8) sin (7 x + 8) + 3 (28 x + 32 - sin (4 (7 x + 8))))

Add a constant to the solution

= \frac{1}{224} (32 cos^{3} (7 x + 8) sin (7 x + 8) + 3 (28 x + 32 - sin (4 (7 x + 8)))) + C

\int ln (10 x) d x

y^{^{'}} - \frac{4}{x} y = \frac{y ^{3}}{x ^{4}}

\int 5 x e^{5 x} d x

\int \frac{1}{x ^{2} - 14 x + 68} d x

\frac{d y}{d x} + (\frac{y}{x}) = 7 x^{3} y^{2}

What is the integral of cos^4(7x+8) ?
The integral of cos^4(7x+8) is 1/224 (32cos^3(7x+8)sin(7x+8)+3(28x+32-sin(4(7x+8))))+C