What is the integral of (tan^3(x))/(sec^6(x)) ?

The integral of (tan^3(x))/(sec^6(x)) is -(cos^4(x))/4+(cos^6(x))/6+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (tan^3(x))/(sec^6(x))

\int \frac{tan ^{3} ( x )}{sec ^{6} ( x )} d x

- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

Solution steps

\int \frac{tan ^{3} ( x )}{sec ^{6} ( x )} d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{3} (x) d x

Apply u-substitution

: \int - u^{3} (1 - u^{2}) d u

= \int - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= \int - u^{3} + u^{5} 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 u^{3} d u + \int u^{5} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = cos (x)

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}

Add a constant to the solution

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

\int 1 - r^{2} r d r

x \to 4 lim (ln (x - 4))

\frac{d}{d x} (\frac{ln ( 1 + x )}{( 1 - x )})

\int_{0}^{3} 4 x - x^{2} d x

\frac{d}{d x} (3 tan (2 x))

What is the integral of (tan^3(x))/(sec^6(x)) ?
The integral of (tan^3(x))/(sec^6(x)) is -(cos^4(x))/4+(cos^6(x))/6+C