What is the integral of (tan^8(x))/(sin^5(x)) ?

The integral of (tan^8(x))/(sin^5(x)) is 1/35 sec^5(x)(-2+5tan^2(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (tan^8(x))/(sin^5(x))

\int \frac{tan ^{8} ( x )}{sin ^{5} ( x )} d x

\frac{1}{35} sec^{5} (x) (- 2 + 5 tan^{2} (x)) + C

Solution steps

\int \frac{tan ^{8} ( x )}{sin ^{5} ( x )} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int (v - 1) v^{\frac{3}{2}} d v

Apply Integration By Parts

: \frac{2}{5} v^{\frac{5}{2}} (v - 1) - \int \frac{2}{5} v^{\frac{5}{2}} d v

= \frac{1}{2} (\frac{2}{5} v^{\frac{5}{2}} (v - 1) - \int \frac{2}{5} v^{\frac{5}{2}} d v)

\int \frac{2}{5} v^{\frac{5}{2}} d v = \frac{4}{35} v^{\frac{7}{2}}

= \frac{1}{2} (\frac{2}{5} v^{\frac{5}{2}} (v - 1) - \frac{4}{35} v^{\frac{7}{2}})

Substitute back

= \frac{1}{2} (\frac{2}{5} (tan^{2} (x) + 1)^{\frac{5}{2}} (tan^{2} (x) + 1 - 1) - \frac{4}{35} (tan^{2} (x) + 1)^{\frac{7}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (tan^{2} (x) + 1)^{\frac{5}{2}} (tan^{2} (x) + 1 - 1) - \frac{4}{35} (tan^{2} (x) + 1)^{\frac{7}{2}}) : \frac{1}{35} sec^{5} (x) (- 2 + 5 tan^{2} (x))

= \frac{1}{35} sec^{5} (x) (- 2 + 5 tan^{2} (x))

Add a constant to the solution

= \frac{1}{35} sec^{5} (x) (- 2 + 5 tan^{2} (x)) + C

\int \frac{tan ( 4 x )}{cos ( 4 x )} d x

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

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

\int \frac{x}{x ^{4} - 13 x ^{2} + 36} d x

\int \frac{20 x}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

What is the integral of (tan^8(x))/(sin^5(x)) ?
The integral of (tan^8(x))/(sin^5(x)) is 1/35 sec^5(x)(-2+5tan^2(x))+C