Integral of tan(x+c)dx dx

The solution

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

Integral(tan(x + c), (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$- \log{\left(\cos{\left(c + x \right)} \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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 | tan(x + c) dx = C - log(cos(x + c))
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$$\int \tan{\left(c + x \right)}\, dx = C - \log{\left(\cos{\left(c + x \right)} \right)}$$

Simplify

The answer [src]

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

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

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

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

log(1 + tan(1 + c)^2)/2 - log(1 + tan(c)^2)/2

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

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Integral of tan(x+c)dx dx

Limits of integration:

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Piecewise:

The solution

Method #1

Method #2