Question

plz
integrate
this question
it will be helpful for me​

Answer 1

We can evaluate the integral without actual integration.

The above expression literally means area under the cosine curve in the interval [π,2π].

We can divide the interval into two, [π,3π/2] and [3π/2,2π]. Also, maxima and minima of the cosine curve are 1 and - 1 (x = 2nπ ± ∅).

Firstly,

$\boxed{\boxed{\displaystyle \sf \int_{\pi}^{2 \pi} cos (\theta)d \theta = \int_{\frac{3 \pi}{2}}^{2 \pi} cos (\theta)d \theta + \int_{\pi}^{\frac{3 \pi}{2}} cos (\theta)d \theta}}$

Cosine curve from π to 3π/2 resembles a triangle (approximation).

$\sf \: Area_1 = \dfrac{1}{2} \times \dfrac{\pi}{2} \times - 1 \\ \\ \longrightarrow \sf \:Area_1 = -\dfrac{\pi}{4} \: sq.units$

Likewise, from 3π/2 to 2π,

$\sf \: Area_2 = \dfrac{1}{2} \times \dfrac{\pi}{2} \times 1 \\ \\ \longrightarrow \sf \:Area_2 = \dfrac{\pi}{4} \: sq.units$

Now, area under the curve will be :

$\longrightarrow \sf \:Area = \dfrac{\pi}{4} \: - \dfrac{\pi}{4} \\ \\ \longrightarrow \boxed{ \boxed{ \sf \:Area = 0 \: sq.units \ (approx)}}$

Thus,

$\boxed{\boxed{\displaystyle \sf \int_{\pi}^{2 \pi} cos (\theta)d \theta = 0}}$