Integral of t³-5 dt

The solution

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    1             
    /             
   |              
   |   / 3    \   
   |   \t  - 5/ dt
   |              
  /               
sin(x)

$$\int\limits_{\sin{\left(x \right)}}^{1} \left(t^{3} - 5\right)\, dt$$

Integral(t^3 - 5, (t, sin(x), 1))

Detail solution

Integrate term-by-term:
1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$ :
  
  $\int t^{3}\, dt = \frac{t^{4}}{4}$
1. The integral of a constant is the constant times the variable of integration:
  
  $\int \left(-5\right)\, dt = - 5 t$
The result is: $\frac{t^{4}}{4} - 5 t$
Now simplify:

$\frac{t \left(t^{3} - 20\right)}{4}$
Add the constant of integration:

$\frac{t \left(t^{3} - 20\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{t \left(t^{3} - 20\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                          
 |                          4
 | / 3    \                t 
 | \t  - 5/ dt = C - 5*t + --
 |                         4 
/

$$\int \left(t^{3} - 5\right)\, dt = C + \frac{t^{4}}{4} - 5 t$$

Simplify

The answer [src]

                     4   
  19              sin (x)
- -- + 5*sin(x) - -------
  4                  4

$$- \frac{\sin^{4}{\left(x \right)}}{4} + 5 \sin{\left(x \right)} - \frac{19}{4}$$

                     4   
  19              sin (x)
- -- + 5*sin(x) - -------
  4                  4

$$- \frac{\sin^{4}{\left(x \right)}}{4} + 5 \sin{\left(x \right)} - \frac{19}{4}$$

-19/4 + 5*sin(x) - sin(x)^4/4

Use the examples entering the upper and lower limits of integration.

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Integral of t³-5 dt

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