What is the integral of 1/2 tcos(4)t^2 ?

The integral of 1/2 tcos(4)t^2 is 1/8 cos(4)t^4+C

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integral of 1/2 tcos(4)t^2

\int \frac{1}{2} t cos (4) t^{2} d t

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

Solution steps

\int \frac{1}{2} t cos (4) t^{2} d t

Take the constant out:

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

= \frac{1}{2} cos (4) \cdot \int t t^{2} d t

Simplify

t t^{2} : t^{3}

= \frac{1}{2} cos (4) \cdot \int t^{3} d t

Apply the Power Rule

: \frac{t ^{4}}{4}

= \frac{1}{2} cos (4) \frac{t ^{4}}{4}

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d y} (y + 1)

\int 6 x + 5 d x

\frac{4}{s} + \frac{6}{s ^{5}} - \frac{1}{s + 8}

y = 2 x^{2} - x, (- 1, 2)

x (ln (3 x))^{2}

What is the integral of 1/2 tcos(4)t^2 ?
The integral of 1/2 tcos(4)t^2 is 1/8 cos(4)t^4+C