What is the integral of 8/(t^9)sin(1/(t^8)-9) ?

The integral of 8/(t^9)sin(1/(t^8)-9) is cos(1/(t^8)-9)+C

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integral of 8/(t^9)sin(1/(t^8)-9)

\int \frac{8}{t ^{9}} sin (\frac{1}{t ^{8}} - 9) d t

cos (\frac{1}{t ^{8}} - 9) + C

Solution steps

\int \frac{8}{t ^{9}} sin (\frac{1}{t ^{8}} - 9) d t

Take the constant out:

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

= 8 \cdot \int \frac{1}{t ^{9}} sin (\frac{1}{t ^{8}} - 9) d t

Apply u-substitution

: \int - \frac{sin ( u )}{8} d u

= 8 \cdot \int - \frac{sin ( u )}{8} d u

Take the constant out:

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

= 8 (- \frac{1}{8} \cdot \int sin (u) d u)

Use the common integral:

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

= 8 (- \frac{1}{8} (- cos (u)))

Substitute back

u = \frac{1}{t ^{8}} - 9

= 8 (- \frac{1}{8} (- cos (\frac{1}{t ^{8}} - 9)))

Simplify

8 (- \frac{1}{8} (- cos (\frac{1}{t ^{8}} - 9))) : cos (\frac{1}{t ^{8}} - 9)

= cos (\frac{1}{t ^{8}} - 9)

Add a constant to the solution

= cos (\frac{1}{t ^{8}} - 9) + C

y^{^{'}} + tan (x) y = cos^{2} (x)

\frac{\partial}{\partial x} (tan (2 x - y))

\frac{\partial}{\partial x} ((x^{2} - 2) e^{x^{2} y})

\int \frac{30}{30 + e ^{x}} d x

y = 4 (6 - x^{2})^{5}

What is the integral of 8/(t^9)sin(1/(t^8)-9) ?
The integral of 8/(t^9)sin(1/(t^8)-9) is cos(1/(t^8)-9)+C