#### To determine

**To evaluate:**

∫cosx1+sinxdx and verify using graphing function

#### Answer

∫cosx1+sinxdx=21+sinx+C

#### Explanation

**1) Concept:**

i) Indefinite integration:

∫fx=F(x) if F'x=f(x)

ii) The Substitution rule:

If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

∫fgxg'(x) dx=∫f(u)du

**2) Formula:**

∫xndx=xn+1n+1+C

**3) Given:**

∫cosx1+sinxdx

**4) Calculation:**

Substitute, u=1+sinx

Differentiating with respect to x,

dudx=cosx

du=cosx dx

∫cosx1+sinxdx=∫duu=∫u-12 du

By using formula,

∫u-12 du=2u12

By plugging value of u,

∫cosx1+sinxdx=21+sinx+C

For, C=0

∫cosx1+sinxdx=21+sinx

Now plot the graph of function f and its antiderivative F’

The graph of both functions is the same on the intervals -π2,3π2

From the graph, it is observed that f is positive from the interval -π2,π2, so F should increase on the interval -π2,π2. At point π2, graph of f changes from positive to negative, so graph of F should be at maximum. Similarly in π2,3π2 f is negative so F must decrease. So from the graph F does have the properties of anti-derivative of f, so the answer is reasonable.

**Conclusion:**

Therefore,

∫cosx1+sinxdx=21+sinx+C