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

The integral of (tan^3(ln(x))sec^6(ln(x)))/x is (tan^4(ln(x)))/4+(tan^6(ln(x)))/3+(tan^8(ln(x)))/8+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(ln(x))sec^6(ln(x)))/x

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

\frac{tan ^{4} ( ln ( x ) )}{4} + \frac{tan ^{6} ( ln ( x ) )}{3} + \frac{tan ^{8} ( ln ( x ) )}{8} + C

Solution steps

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

Apply u-substitution

: \int sec^{6} (u) tan^{3} (u) d u

= \int sec^{6} (u) tan^{3} (u) d u

Rewrite using trig identities

= \int sec^{2} (u) (1 + tan^{2} (u))^{2} tan^{3} (u) d u

Apply u-substitution

: \int v^{3} (1 + v^{2})^{2} d v

= \int v^{3} (1 + v^{2})^{2} d v

Expand

v^{3} (1 + v^{2})^{2} : v^{3} + 2 v^{5} + v^{7}

= \int v^{3} + 2 v^{5} + v^{7} d v

Apply the Sum Rule:

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

= \int v^{3} d v + \int 2 v^{5} d v + \int v^{7} d v

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

\int 2 v^{5} d v = \frac{v ^{6}}{3}

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{v ^{4}}{4} + \frac{v ^{6}}{3} + \frac{v ^{8}}{8}

Substitute back

= \frac{tan ^{4} ( ln ( x ) )}{4} + \frac{tan ^{6} ( ln ( x ) )}{3} + \frac{tan ^{8} ( ln ( x ) )}{8}

Add a constant to the solution

= \frac{tan ^{4} ( ln ( x ) )}{4} + \frac{tan ^{6} ( ln ( x ) )}{3} + \frac{tan ^{8} ( ln ( x ) )}{8} + C

\int \frac{5 x - 7}{2 x ^{2} - 3 x + 7} d x

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

\int 2 t e^{t^{2}} d t

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

\int \frac{x + 1}{x ^{2} - 6 x + 8} d x

What is the integral of (tan^3(ln(x))sec^6(ln(x)))/x ?
The integral of (tan^3(ln(x))sec^6(ln(x)))/x is (tan^4(ln(x)))/4+(tan^6(ln(x)))/3+(tan^8(ln(x)))/8+C