What is the integral of (sin^3(x))/(cos^2(x)\sqrt[3]{cos(x))} ?

The integral of (sin^3(x))/(cos^2(x)\sqrt[3]{cos(x))} is -(-3/4 sec^{4/3}(x)-3/2 cos^{2/3}(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin^3(x))/(cos^2(x)\sqrt[3]{cos(x))}

- (- \frac{3}{4} sec^{\frac{4}{3}} (x) - \frac{3}{2} cos^{\frac{2}{3}} (x)) + C

Solution steps

Apply u-substitution

Take the constant out:

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

Simplify

= - \int \frac{1 - u ^{2}}{u ^{\frac{7}{3}}} d u

Expand

\frac{1 - u ^{2}}{u ^{\frac{7}{3}}} : \frac{1}{u ^{\frac{7}{3}}} - \frac{1}{u ^{\frac{1}{3}}}

= - \int \frac{1}{u ^{\frac{7}{3}}} - \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Sum Rule:

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

= - (\int \frac{1}{u ^{\frac{7}{3}}} d u - \int \frac{1}{u ^{\frac{1}{3}}} d u)

\int \frac{1}{u ^{\frac{7}{3}}} d u = - \frac{3}{4 u ^{\frac{4}{3}}}

\int \frac{1}{u ^{\frac{1}{3}}} d u = \frac{3}{2} u^{\frac{2}{3}}

= - (- \frac{3}{4 u ^{\frac{4}{3}}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = cos (x)

= - (- \frac{3}{4 cos ^{\frac{4}{3}} ( x )} - \frac{3}{2} cos^{\frac{2}{3}} (x))

Simplify

- (- \frac{3}{4 cos ^{\frac{4}{3}} ( x )} - \frac{3}{2} cos^{\frac{2}{3}} (x)) : - (- \frac{3}{4} sec^{\frac{4}{3}} (x) - \frac{3}{2} cos^{\frac{2}{3}} (x))

= - (- \frac{3}{4} sec^{\frac{4}{3}} (x) - \frac{3}{2} cos^{\frac{2}{3}} (x))

Add a constant to the solution

= - (- \frac{3}{4} sec^{\frac{4}{3}} (x) - \frac{3}{2} cos^{\frac{2}{3}} (x)) + C

(- 5, 3), (5, - 5)

\frac{d}{d x} (x cos (θ))

\int x y e^{- y x} d x

x^{2} + x y - y^{2} = - 11, (1, 4)

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

What is the integral of (sin^3(x))/(cos^2(x)\sqrt[3]{cos(x))} ?
The integral of (sin^3(x))/(cos^2(x)\sqrt[3]{cos(x))} is -(-3/4 sec^{4/3}(x)-3/2 cos^{2/3}(x))+C