English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to pi/6 of cos^{-3}(2x)sin(2x)

الحلّ

\int_{0}^{\frac{π}{6}} cos^{- 3} (2 x) sin (2 x) d x

\frac{3}{4}

عشري

0.75

خطوات الحلّ

\int_{0}^{\frac{π}{6}} cos^{- 3} (2 x) sin (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} \frac{2 cos ( x ) sin ( x )}{cos ^{3} ( 2 x )} d x

\int a \cdot f (x) d x = a \cdot \int f (x) d x

:قم بإخراج الثابت

= 2 \cdot \int_{0}^{\frac{π}{6}} \frac{cos ( x ) sin ( x )}{cos ^{3} ( 2 x )} d x

Rewrite using trig identities

= 2 \cdot \int_{0}^{\frac{π}{6}} \frac{sin ( 2 x )}{2 cos ^{3} ( 2 x )} d x

\int a \cdot f (x) d x = a \cdot \int f (x) d x

:قم بإخراج الثابت

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

فعّل طريقة التكامل بالتعويض

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} \frac{sin ( u )}{2 cos ^{3} ( u )} d u

\int a \cdot f (x) d x = a \cdot \int f (x) d x

:قم بإخراج الثابت

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} \frac{sin ( u )}{cos ^{3} ( u )} d u

Rewrite using trig identities

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} tan (u) sec^{2} (u) d u

فعّل طريقة التكامل بالتعويض

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{3} v d v

استعمل قانون الأسس

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

2 \cdot \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

بسّط:

\frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

= \frac{1}{2} [\frac{v ^{2}}{2}]_{0}^{3}

احسب التكامل المحدّد:

\frac{3}{2}

= \frac{1}{2} \cdot \frac{3}{2}

بسّط

= \frac{3}{4}

\int_{1}^{e} \frac{ln ( x ^{2} )}{x} d x

\int_{0}^{2 π} 1 - cos (x) d x

\int_{1}^{3} 4 x - x^{2} d x

\int_{0}^{x} \frac{sin ( x )}{x} d x

\int_{e^{y}}^{y} \frac{y ln ( x )}{x} d x