Question

How do you evaluate the definite integral `int (sinx+cosx)^2dx` from `[pi/6,pi/4]`?

Answer 1

`int_(pi/6)^(pi/4)(sinx + cosx)^2dx = (pi + 3)/12 ~~ 0.51`

Explanation:

Expand:

`=int_(pi/6)^(pi/4)(sin^2x + cos^2x + 2sinxcosx)dx`

Apply the identity `sin^2x + cos^2x = 1`:

`=int_(pi/6)^(pi/4) (1 + 2sinxcosx)dx`

Apply the identity `sin2x = 2sinxcosx`:

`=int_(pi/6)^(pi/4) (1 + sin2x)dx`

Let `u= 2x`. Then `du = 2dx` and `dx= 1/2du`.

`=int_(pi/6)^(pi/4) (1 + sinu) * 1/2du`

Use the property `int(af(x)) = aintf(x)` where `a` is a constant.

`=1/2int_(pi/6)^(pi/4) (1 + sinu)du`

Make sure you know your basic integrals of trig functions (e.g. sine and cosine). Also, it would be good to commit to memory the formula `int(x^n)dx = x^(n + 1)/(n + 1) + C`. `color(red)("Remember")`: we are integrating with respect to `u`.

`=1/2[u - cosu]_(pi/6)^(pi/4)`

Since `u = 2x`:

`=1/2[2x - cos(2x)]_(pi/6)^(pi/4)`

Evaluate using `int_a^bF(x) = f(b) - f(a)`, where `f'(x) = F(x)`.

`=1/2[2(pi/4) - cos(2(pi/4))] - 1/2[2(pi/6) - cos(2(pi/6))]`

`=1/2[pi/2 - 0] - 1/2[pi/3 - 1/2]`

`=pi/4 - pi/6 + 1/4`

`= (pi + 3)/12`

`~~0.51`

Hopefully this helps!