Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^(-1/2)
Integral of 1/sqrt(1+u^2)
Integral of x^4lnx
Integral of (2x+3)
Identical expressions
five sin2x/(cos2x)^5
5 sinus of 2x divide by ( co sinus of e of 2x) to the power of 5
five sinus of 2x divide by ( co sinus of e of 2x) to the power of 5
5sin2x/(cos2x)5
5sin2x/cos2x5
5sin2x/(cos2x)⁵
5sin2x/cos2x^5
5sin2x divide by (cos2x)^5
5sin2x/(cos2x)^5dx

Integral of 5sin2x/(cos2x)^5 dx

The solution

You have entered [src]

  1              
  /              
 |               
 |  5*sin(2*x)   
 |  ---------- dx
 |     5         
 |  cos (2*x)    
 |               
/                
0

$$\int\limits_{0}^{1} \frac{5 \sin{\left(2 x \right)}}{\cos^{5}{\left(2 x \right)}}\, dx$$

Integral((5*sin(2*x))/cos(2*x)^5, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $\frac{5 du}{2}$ :
  
  $\int \frac{5 \sin{\left(u \right)}}{2 \cos^{5}{\left(u \right)}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{\cos^{5}{\left(u \right)}}\, du = \frac{5 \int \frac{\sin{\left(u \right)}}{\cos^{5}{\left(u \right)}}\, du}{2}$
    1. Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u^{5}}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u^{5}}\, du = - \int \frac{1}{u^{5}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{5}}\, du = - \frac{1}{4 u^{4}}$
      
      So, the result is: $\frac{1}{4 u^{4}}$
      
      Now substitute $u$ back in:
      
      $\frac{1}{4 \cos^{4}{\left(u \right)}}$
    So, the result is: $\frac{5}{8 \cos^{4}{\left(u \right)}}$
  Now substitute $u$ back in:
  
  $\frac{5}{8 \cos^{4}{\left(2 x \right)}}$
Method #2
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{10 \sin{\left(x \right)} \cos{\left(x \right)}}{\cos^{5}{\left(2 x \right)}}\, dx = 10 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\cos^{5}{\left(2 x \right)}}\, dx$
  1. Rewrite the integrand:
    
    $\frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\cos^{5}{\left(2 x \right)}} = \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{32 \cos^{10}{\left(x \right)} - 80 \cos^{8}{\left(x \right)} + 80 \cos^{6}{\left(x \right)} - 40 \cos^{4}{\left(x \right)} + 10 \cos^{2}{\left(x \right)} - 1}$
  2. Let $u = \cos^{2}{\left(x \right)}$ .
    
    Then let $du = - 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \left(- \frac{1}{64 u^{5} - 160 u^{4} + 160 u^{3} - 80 u^{2} + 20 u - 2}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{64 u^{5} - 160 u^{4} + 160 u^{3} - 80 u^{2} + 20 u - 2}\, du = - \int \frac{1}{64 u^{5} - 160 u^{4} + 160 u^{3} - 80 u^{2} + 20 u - 2}\, du$
      
      Rewrite the integrand:
      
      $\frac{1}{64 u^{5} - 160 u^{4} + 160 u^{3} - 80 u^{2} + 20 u - 2} = \frac{1}{2 \left(2 u - 1\right)^{5}}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(2 u - 1\right)^{5}}\, du = \frac{\int \frac{1}{\left(2 u - 1\right)^{5}}\, du}{2}$
      
      Let $u = 2 u - 1$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u^{5}}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u^{5}}\, du = \frac{\int \frac{1}{u^{5}}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{5}}\, du = - \frac{1}{4 u^{4}}$
      
      So, the result is: $- \frac{1}{8 u^{4}}$
      
      Now substitute $u$ back in:
      
      $- \frac{1}{8 \left(2 u - 1\right)^{4}}$
      
      So, the result is: $- \frac{1}{16 \left(2 u - 1\right)^{4}}$
      
      So, the result is: $\frac{1}{16 \left(2 u - 1\right)^{4}}$
    Now substitute $u$ back in:
    
    $\frac{1}{16 \left(2 \cos^{2}{\left(x \right)} - 1\right)^{4}}$
  So, the result is: $\frac{5}{8 \left(2 \cos^{2}{\left(x \right)} - 1\right)^{4}}$
Add the constant of integration:

$\frac{5}{8 \cos^{4}{\left(2 x \right)}}+ \mathrm{constant}$

The answer is:

$\frac{5}{8 \cos^{4}{\left(2 x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                               
 |                                
 | 5*sin(2*x)               5     
 | ---------- dx = C + -----------
 |    5                     4     
 | cos (2*x)           8*cos (2*x)
 |                                
/

$$\int \frac{5 \sin{\left(2 x \right)}}{\cos^{5}{\left(2 x \right)}}\, dx = C + \frac{5}{8 \cos^{4}{\left(2 x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$\text{NaN}$$

nan

$$\text{NaN}$$

nan

Numerical answer [src]

-17560707880.4313

-17560707880.4313

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

Other calculators

How to use it?

Identical expressions

Integral of 5sin2x/(cos2x)^5 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2