What is the integral from 0 to pi/2 of 1+cos(2x) ?

The integral from 0 to pi/2 of 1+cos(2x) is pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of 1+cos(2x)

\int_{0}^{\frac{π}{2}} 1 + cos (2 x) d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 1 + cos (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{π}{2}} 1 d x + \int_{0}^{\frac{π}{2}} cos (2 x) d x

\int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 x) d x = 0

= \frac{π}{2} + 0

Simplify

= \frac{π}{2}

\int_{2}^{3} (- 16 x + 2) d x

\int_{\frac{π}{2}}^{2 π} 2 sin (x) d x

\int_{- 7}^{7} ∣ 6 x ∣ - (x^{2} - 7) d x

\int_{0}^{1} π (- x^{2} + 1)^{2} d x

\int_{- 1}^{1} x^{1} d x

What is the integral from 0 to pi/2 of 1+cos(2x) ?
The integral from 0 to pi/2 of 1+cos(2x) is pi/2