What is the integral of (t^2cos(2t^3))/(csc(10t^3)) ?

The integral of (t^2cos(2t^3))/(csc(10t^3)) is 1/6 (-1/12 cos(12t^3)-1/8 cos(8t^3))+C

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integral of (t^2cos(2t^3))/(csc(10t^3))

\int \frac{t ^{2} cos ( 2 t ^{3} )}{csc ( 10 t ^{3} )} d t

\frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) + C

Solution steps

\int \frac{t ^{2} cos ( 2 t ^{3} )}{csc ( 10 t ^{3} )} d t

Apply u-substitution

= \int \frac{cos ( 2 u )}{3 csc ( 10 u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{cos ( 2 u )}{csc ( 10 u )} d u

Use the basic trigonometric identity:

\frac{1}{csc ( x )} = sin (x)

= \frac{1}{3} \cdot \int cos (2 u) sin (10 u) d u

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \frac{1}{3} \cdot \int \frac{sin ( 10 u + 2 u ) + sin ( 10 u - 2 u )}{2} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int sin (10 u + 2 u) + sin (10 u - 2 u) d u

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}{3} \cdot \frac{1}{2} (\int sin (10 u + 2 u) d u + \int sin (10 u - 2 u) d u)

\int sin (10 u + 2 u) d u = - \frac{1}{12} cos (12 u)

\int sin (10 u - 2 u) d u = - \frac{1}{8} cos (8 u)

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 u) - \frac{1}{8} cos (8 u))

Substitute back

u = t^{3}

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

Simplify

\frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) : \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

= \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3}))

Add a constant to the solution

= \frac{1}{6} (- \frac{1}{12} cos (12 t^{3}) - \frac{1}{8} cos (8 t^{3})) + C

What is the integral of (t^2cos(2t^3))/(csc(10t^3)) ?
The integral of (t^2cos(2t^3))/(csc(10t^3)) is 1/6 (-1/12 cos(12t^3)-1/8 cos(8t^3))+C