What is the integral of tan^2(x^5)x^4 ?

The integral of tan^2(x^5)x^4 is 1/5 (-x^5+tan(x^5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^2(x^5)x^4

\int tan^{2} (x^{5}) x^{4} d x

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

Solution steps

\int tan^{2} (x^{5}) x^{4} d x

Apply u-substitution

: \int \frac{tan ^{2} ( u )}{5} d u

= \int \frac{tan ^{2} ( u )}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int - 1 + sec^{2} (u) 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

= \frac{1}{5} (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= \frac{1}{5} (- u + tan (u))

Substitute back

u = x^{5}

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} + \frac{1}{4}} d x

\int \frac{1}{9} cos (x) d x

\int 10 x (5 x^{2} + 9)^{7} d x

\int x^{\frac{2}{6}} d x

What is the integral of tan^2(x^5)x^4 ?
The integral of tan^2(x^5)x^4 is 1/5 (-x^5+tan(x^5))+C