Question

intreal 0 to 2π cos^2 theta * sin theta dtheta​

Answer 1

Answer:

0

Step-by-step explanation:

$\\\tt \int\limits^{2\pi}_0 {cos^2 \theta \ sin\theta} \, d\theta$

Let y= cosθ

⇒ dy = - sinθ dθ

Now the values of limits will become 1 to 1

Since when x= 0, y = 1 and when x= 2π, y =1

$\\\tt \int\limits^{1}_1 {y^2 \ sin\theta} \, \dfrac{dy}{-sin \theta} \\\\\implies \int\limits^{1}_1 {-y^2 } \, dy \\ \implies [\dfrac{-y^3}{3} ] _{1} ^{1} \\=\dfrac{-1}{3} - (\dfrac{-1}{3} ) \\=0$