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