What is the integral of 9t^{11}cos(1-t^6) ?

The integral of 9t^{11}cos(1-t^6) is -3/2 (t^6sin(-t^6+1)-cos(-t^6+1))+C

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integral of 9t^{11}cos(1-t^6)

\int 9 t^{11} cos (1 - t^{6}) d t

- \frac{3}{2} (t^{6} sin (- t^{6} + 1) - cos (- t^{6} + 1)) + C

Solution steps

\int 9 t^{11} cos (1 - t^{6}) d t

Take the constant out:

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

= 9 \cdot \int t^{11} cos (1 - t^{6}) d t

Apply u-substitution

: \int - \frac{( - u + 1 ) cos ( u )}{6} d u

= 9 \cdot \int - \frac{( - u + 1 ) cos ( u )}{6} d u

Take the constant out:

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

= 9 (- \frac{1}{6} \cdot \int (- u + 1) cos (u) d u)

Apply Integration By Parts

: sin (u) (- u + 1) - \int - sin (u) d u

= 9 (- \frac{1}{6} (sin (u) (- u + 1) - \int - sin (u) d u))

Take the constant out:

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

= 9 (- \frac{1}{6} (sin (u) (- u + 1) - (- \int sin (u) d u)))

Use the common integral:

\int sin (u) d u = - cos (u)

= 9 (- \frac{1}{6} (sin (u) (- u + 1) - (- (- cos (u)))))

Substitute back

u = 1 - t^{6}

= 9 (- \frac{1}{6} (sin (1 - t^{6}) (- (1 - t^{6}) + 1) - (- (- cos (1 - t^{6})))))

Simplify

9 (- \frac{1}{6} (sin (1 - t^{6}) (- (1 - t^{6}) + 1) - (- (- cos (1 - t^{6}))))) : - \frac{3}{2} (t^{6} sin (- t^{6} + 1) - cos (- t^{6} + 1))

= - \frac{3}{2} (t^{6} sin (- t^{6} + 1) - cos (- t^{6} + 1))

Add a constant to the solution

= - \frac{3}{2} (t^{6} sin (- t^{6} + 1) - cos (- t^{6} + 1)) + C

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\int 85 - 3 25 - x d x

\int \frac{u}{1} d u

\frac{d}{d x} (6 x^{- 2} + 9 x^{- 1})

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

What is the integral of 9t^{11}cos(1-t^6) ?
The integral of 9t^{11}cos(1-t^6) is -3/2 (t^6sin(-t^6+1)-cos(-t^6+1))+C