What is the integral of 9cos^4(x) ?

The integral of 9cos^4(x) is 9(cos^3(x)sin(x)+3/8 (x-1/4 sin(4x)))+C

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integral of 9cos^4(x)

\int 9 cos^{4} (x) d x

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

Solution steps

\int 9 cos^{4} (x) d x

Take the constant out:

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

= 9 \cdot \int cos^{4} (x) d x

Simplify

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

= 9 \cdot \int cos^{3} (x) cos (x) d x

Apply Integration By Parts

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

= 9 (cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x)

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

= 9 (cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x))))

Simplify

= 9 (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)))

Add a constant to the solution

= 9 (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + C

(2, 4), (3, 6)

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

y = e^{1 - x}

\frac{d}{d x} (ln^{2} (3 x))

F (s) = (\frac{2}{s} - \frac{1}{s ^{3}})^{2}

What is the integral of 9cos^4(x) ?
The integral of 9cos^4(x) is 9(cos^3(x)sin(x)+3/8 (x-1/4 sin(4x)))+C