Integral of sin(x+y) dx

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

Integral(sin(x + y), (x, 0, 1))

Detail solution

Let $u = x + y$ .

Then let $du = dx$ and substitute $du$ :

$\int \sin{\left(u \right)}\, du$
1. The integral of sine is negative cosine:
  
  $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
Now substitute $u$ back in:

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

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

The answer is:

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

The answer (Indefinite) [src]

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$$\int \sin{\left(x + y \right)}\, dx = C - \cos{\left(x + y \right)}$$

Simplify

The answer [src]

-cos(1 + y) + cos(y)

$$\cos{\left(y \right)} - \cos{\left(y + 1 \right)}$$

-cos(1 + y) + cos(y)

$$\cos{\left(y \right)} - \cos{\left(y + 1 \right)}$$

-cos(1 + y) + cos(y)

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