What is the integral of tcos^5(t^2) ?

The integral of tcos^5(t^2) is 1/2 sin(t^2)-1/3 sin^3(t^2)+1/10 sin^5(t^2)+C

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integral of tcos^5(t^2)

\int t cos^{5} (t^{2}) d t

\frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2}) + C

Solution steps

\int t cos^{5} (t^{2}) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int (1 - v^{2})^{2} d v

= \frac{1}{2} \cdot \int (1 - v^{2})^{2} d v

Expand

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= \frac{1}{2} \cdot \int 1 - 2 v^{2} + v^{4} d v

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}{2} (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{2} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{2} (sin (t^{2}) - \frac{2 sin ^{3} ( t ^{2} )}{3} + \frac{sin ^{5} ( t ^{2} )}{5})

Simplify

\frac{1}{2} (sin (t^{2}) - \frac{2 sin ^{3} ( t ^{2} )}{3} + \frac{sin ^{5} ( t ^{2} )}{5}) : \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2})

= \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2})

Add a constant to the solution

= \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2}) + C

y = 1 + 2 e^{8 x}

n = 1 \sum \infty \frac{1}{7 n ^{2} + n + 6}

\frac{d}{d x} (- c e^{- x})

\int \frac{1}{( 2 x - 3 ) ( 3 x ^{2} + x )} d x

\int 2 x^{5} e^{x} d x

What is the integral of tcos^5(t^2) ?
The integral of tcos^5(t^2) is 1/2 sin(t^2)-1/3 sin^3(t^2)+1/10 sin^5(t^2)+C