Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^4/(1+x^2)
Integral of (x²-1)/(x²+1)
Integral of x^2/(√(1-x^2))
Integral of x^2/(1+x^2)^2
Identical expressions
48cos^4t*sin^2t
48 co sinus of e of to the power of 4t multiply by sinus of squared t
48cos4t*sin2t
48cos⁴t*sin²t
48cos to the power of 4t*sin to the power of 2t
48cos^4tsin^2t
48cos4tsin2t

Solving integrals/ 48cos^4t*sin^2t

Integral of 48cos^4t*sin^2t dt

The solution

You have entered [src]

 pi                      
 --                      
 2                       
  /                      
 |                       
 |        4       2      
 |  48*cos (t)*sin (t) dt
 |                       
/                        
pi                       
--                       
6

$$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} 48 \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}\, dt$$

Integral(48*cos(t)^4*sin(t)^2, (t, pi/6, pi/2))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 48 \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}\, dt = 48 \int \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}\, dt$
1. Rewrite the integrand:
  
  $\sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)^{2}$
2. There are multiple ways to do this integral.
  Method #1
  1. Let $u = 2 t$ .
    
    Then let $du = 2 dt$ and substitute $du$ :
    
    $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} + \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
    1. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{3}$
      
      The result is: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin^{3}{\left(u \right)}}{48} - \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du = \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
      
      The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} + \frac{\sin^{3}{\left(u \right)}}{48}$
    Now substitute $u$ back in:
    
    $\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
  Method #2
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)^{2} = - \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{3}{\left(2 t \right)}\, dt}{8}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 t \right)} = \left(1 - \sin^{2}{\left(2 t \right)}\right) \cos{\left(2 t \right)}$
      
      Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 t \right)}}{6} + \frac{\sin{\left(2 t \right)}}{2}$
      
      So, the result is: $\frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}$
      
      Let $u = 4 t$ .
      
      Then let $du = 4 dt$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 t \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 t \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dt = \frac{t}{2}$
      
      The result is: $\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}$
      
      So, the result is: $- \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 t \right)}}{8}\, dt = \frac{\int \cos{\left(2 t \right)}\, dt}{8}$
      
      Let $u = 2 t$ .
      
      Then let $du = 2 dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 t \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 t \right)}}{16}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dt = \frac{t}{8}$
    The result is: $\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
  Method #3
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right) \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)^{2} = - \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} + \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{3}{\left(2 t \right)}\, dt}{8}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 t \right)} = \left(1 - \sin^{2}{\left(2 t \right)}\right) \cos{\left(2 t \right)}$
      
      Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 t \right)}}{6} + \frac{\sin{\left(2 t \right)}}{2}$
      
      So, the result is: $\frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(2 t \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}$
      
      Let $u = 4 t$ .
      
      Then let $du = 4 dt$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 t \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 t \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dt = \frac{t}{2}$
      
      The result is: $\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}$
      
      So, the result is: $- \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 t \right)}}{8}\, dt = \frac{\int \cos{\left(2 t \right)}\, dt}{8}$
      
      Let $u = 2 t$ .
      
      Then let $du = 2 dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 t \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 t \right)}}{16}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dt = \frac{t}{8}$
    The result is: $\frac{t}{16} + \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
So, the result is: $3 t + \sin^{3}{\left(2 t \right)} - \frac{3 \sin{\left(4 t \right)}}{4}$
Add the constant of integration:

$3 t + \sin^{3}{\left(2 t \right)} - \frac{3 \sin{\left(4 t \right)}}{4}+ \mathrm{constant}$

The answer is:

$3 t + \sin^{3}{\left(2 t \right)} - \frac{3 \sin{\left(4 t \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                        
 |                                                         
 |       4       2                3              3*sin(4*t)
 | 48*cos (t)*sin (t) dt = C + sin (2*t) + 3*t - ----------
 |                                                   4     
/

$$6\,\left({{2\,t-{{\sin \left(4\,t\right)}\over{2}}}\over{4}}+{{ \sin ^3\left(2\,t\right)}\over{6}}\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

pi

$$48\,\left(-{{3\,\sin \left(2\,\pi\right)-4\,\sin ^3\pi-6\,\pi }\over{192}}-{{2\,\pi-3\,\sin \left({{2\,\pi}\over{3}}\right)+4\, \sin ^3\left({{\pi}\over{3}}\right)}\over{192}}\right)$$

pi

$$\pi$$

Simplify

Numerical answer [src]

3.14159265358979

3.14159265358979

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of 48cos^4t*sin^2t dt

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3