What is the integral of 2/(t^3)sin(1/(t^2)-4) ?

The integral of 2/(t^3)sin(1/(t^2)-4) is cos(1/(t^2)-4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(t^3)sin(1/(t^2)-4)

\int \frac{2}{t ^{3}} sin (\frac{1}{t ^{2}} - 4) d t

cos (\frac{1}{t ^{2}} - 4) + C

Solution steps

\int \frac{2}{t ^{3}} sin (\frac{1}{t ^{2}} - 4) d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{t ^{3}} sin (\frac{1}{t ^{2}} - 4) d t

Apply u-substitution

: \int - \frac{sin ( u )}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

= 2 (- \frac{1}{2} (- cos (u)))

Substitute back

u = \frac{1}{t ^{2}} - 4

= 2 (- \frac{1}{2} (- cos (\frac{1}{t ^{2}} - 4)))

Simplify

2 (- \frac{1}{2} (- cos (\frac{1}{t ^{2}} - 4))) : cos (\frac{1}{t ^{2}} - 4)

= cos (\frac{1}{t ^{2}} - 4)

Add a constant to the solution

= cos (\frac{1}{t ^{2}} - 4) + C

\int 5 x^{8} d x

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

\int cos (x n) d x

\int 6^{x + 1} d x

\int \frac{2 x ^{2} + 4 x}{2 x} d x

What is the integral of 2/(t^3)sin(1/(t^2)-4) ?
The integral of 2/(t^3)sin(1/(t^2)-4) is cos(1/(t^2)-4)+C