What is area 1,cos^2(x),[0,pi] ?

The answer to area 1,cos^2(x),[0,pi] is 1/2 ((1*0,pi-1)+1/2 (sin(2[0,pi])-sin(2)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 1,cos^2(x),[0,pi]

area

1, cos^{2} (x), [0, π]

\frac{1}{2} ((1 \cdot 0, π - 1) + \frac{1}{2} (sin (2 [0, π]) - sin (2)))

Solution steps

Apply the area formula:

\int_{1}^{[0, π]} ∣ cos (x) ∣^{2} d x

Solve

\int_{1}^{[0, π]} ∣ cos (x) ∣^{2} d x : \frac{1}{2} ((1 \cdot 0, π - 1) + \frac{1}{2} (sin (2 [0, π]) - sin (2)))

The area is:

= \frac{1}{2} ((1 \cdot 0, π - 1) + \frac{1}{2} (sin (2 [0, π]) - sin (2)))

f (5) = \frac{4 x ^{2} + 7 x + 7}{x}

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{2}})

\int sin^{3} (2 θ) cos (2 θ) d θ

\int (4 sin (x)) d x

\frac{\partial}{\partial x} (2 sin^{2} (x))

What is area 1,cos^2(x),[0,pi] ?
The answer to area 1,cos^2(x),[0,pi] is 1/2 ((1*0,pi-1)+1/2 (sin(2[0,pi])-sin(2)))