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Similar expressions
1/sin^2x-4cos^2x

Integral of 1/sin^2x+4cos^2x dx

The solution

You have entered [src]

 pi                         
 --                         
 4                          
  /                         
 |                          
 |  /   1           2   \   
 |  |------- + 4*cos (x)| dx
 |  |   2               |   
 |  \sin (x)            /   
 |                          
/                           
0

$$\int\limits_{0}^{\frac{\pi}{4}} \left(4 \cos^{2}{\left(x \right)} + \frac{1}{\sin^{2}{\left(x \right)}}\right)\, dx$$

Integral(1/(sin(x)^2) + 4*cos(x)^2, (x, 0, pi/4))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 4 \cos^{2}{\left(x \right)}\, dx = 4 \int \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
  So, the result is: $2 x + \sin{\left(2 x \right)}$
1. Don't know the steps in finding this integral.
  
  But the integral is
  
  $- \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$
The result is: $2 x + \sin{\left(2 x \right)} - \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$
Now simplify:

$2 x + \sin{\left(2 x \right)} - \frac{1}{\tan{\left(x \right)}}$
Add the constant of integration:

$2 x + \sin{\left(2 x \right)} - \frac{1}{\tan{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$2 x + \sin{\left(2 x \right)} - \frac{1}{\tan{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      
 |                                                       
 | /   1           2   \                cos(x)           
 | |------- + 4*cos (x)| dx = C + 2*x - ------ + sin(2*x)
 | |   2               |                sin(x)           
 | \sin (x)            /                                 
 |                                                       
/

$$\int \left(4 \cos^{2}{\left(x \right)} + \frac{1}{\sin^{2}{\left(x \right)}}\right)\, dx = C + 2 x + \sin{\left(2 x \right)} - \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

Numerical answer [src]

1.75598925958649e+19

1.75598925958649e+19

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

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Integral of 1/sin^2x+4cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution