What is the integral of tcsc^2((3t^2)/2) ?

The integral of tcsc^2((3t^2)/2) is -1/3 cot((3t^2)/2)+C

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integral of tcsc^2((3t^2)/2)

\int t csc^{2} (\frac{3 t ^{2}}{2}) d t

- \frac{1}{3} cot (\frac{3 t ^{2}}{2}) + C

Solution steps

\int t csc^{2} (\frac{3 t ^{2}}{2}) d t

Apply u-substitution

: \int \frac{csc ^{2} ( \frac{3 u}{2} )}{2} d u

= \int \frac{csc ^{2} ( \frac{3 u}{2} )}{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 csc^{2} (\frac{3 u}{2}) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{1}{sin ^{2} ( \frac{3 u}{2} )} d u

Apply u-substitution

: \int \frac{2}{3 sin ^{2} ( v )} d v

= \frac{1}{2} \cdot \int \frac{2}{3 sin ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{2}{3} \cdot \int \frac{1}{sin ^{2} ( v )} d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{2}{3} (- cot (v))

Substitute back

= \frac{1}{2} \cdot \frac{2}{3} (- cot (\frac{3 t ^{2}}{2}))

Simplify

\frac{1}{2} \cdot \frac{2}{3} (- cot (\frac{3 t ^{2}}{2})) : - \frac{1}{3} cot (\frac{3 t ^{2}}{2})

= - \frac{1}{3} cot (\frac{3 t ^{2}}{2})

Add a constant to the solution

= - \frac{1}{3} cot (\frac{3 t ^{2}}{2}) + C

n = 4 \sum \infty \frac{1}{n ^{2}}

\int sin (10 a) b xx d x

f (x) = \frac{6}{5 x + 2}, (- 1, - 2)

\int (\frac{sin ( 4 x )}{4}) - (\frac{sin ^{3} ( x )}{4}) d x

n = 0 \sum \infty n (n + 1) - 1

What is the integral of tcsc^2((3t^2)/2) ?
The integral of tcsc^2((3t^2)/2) is -1/3 cot((3t^2)/2)+C