Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of 2*x/(-1+x)
Integral of (x+1)e^(2x)
Integral of (x+1)/(2x²+3x-4)
Integral of ln(x+√(1+x^2))
Derivative of:
(1+tanx)^2
Identical expressions
(one +tanx)^ two
(1 plus tangent of x) squared
(one plus tangent of x) to the power of two
(1+tanx)2
1+tanx2
(1+tanx)²
(1+tanx) to the power of 2
1+tanx^2
(1+tanx)^2dx
Similar expressions
(1-tanx)^2

Solving integrals/ (1+tanx)^2

Integral of (1+tanx)^2 dx

The solution

You have entered [src]

  1                 
  /                 
 |                  
 |              2   
 |  (1 + tan(x))  dx
 |                  
/                   
0

$$\int\limits_{0}^{1} \left(\tan{\left(x \right)} + 1\right)^{2}\, dx$$

Integral((1 + tan(x))^2, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\tan{\left(x \right)} + 1\right)^{2} = \tan^{2}{\left(x \right)} + 2 \tan{\left(x \right)} + 1$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$
  2. Integrate term-by-term:
    1. $\int \sec^{2}{\left(x \right)}\, dx = \tan{\left(x \right)}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
    The result is: $- x + \tan{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \tan{\left(x \right)}\, dx = 2 \int \tan{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
    2. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
    So, the result is: $- 2 \log{\left(\cos{\left(x \right)} \right)}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  The result is: $- 2 \log{\left(\cos{\left(x \right)} \right)} + \tan{\left(x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\left(\tan{\left(x \right)} + 1\right)^{2} = \tan^{2}{\left(x \right)} + 2 \tan{\left(x \right)} + 1$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$
  2. Integrate term-by-term:
    1. $\int \sec^{2}{\left(x \right)}\, dx = \tan{\left(x \right)}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, dx = - x$
    The result is: $- x + \tan{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \tan{\left(x \right)}\, dx = 2 \int \tan{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
    2. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
    So, the result is: $- 2 \log{\left(\cos{\left(x \right)} \right)}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  The result is: $- 2 \log{\left(\cos{\left(x \right)} \right)} + \tan{\left(x \right)}$
Add the constant of integration:

$- 2 \log{\left(\cos{\left(x \right)} \right)} + \tan{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- 2 \log{\left(\cos{\left(x \right)} \right)} + \tan{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                              
 |             2                                
 | (1 + tan(x))  dx = C - 2*log(cos(x)) + tan(x)
 |                                              
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

   /       2   \         
log\1 + tan (1)/ + tan(1)

$$\log{\left(1 + \tan^{2}{\left(1 \right)} \right)} + \tan{\left(1 \right)}$$

   /       2   \         
log\1 + tan (1)/ + tan(1)

$$\log{\left(1 + \tan^{2}{\left(1 \right)} \right)} + \tan{\left(1 \right)}$$

log(1 + tan(1)^2) + tan(1)

Numerical answer [src]

2.78866066542693

2.78866066542693

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (1+tanx)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2