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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x sec^{2} (x^{2}) tan^{4} (x^{2}) d x

Apply u-substitution

: \int \frac{tan ^{4} ( u ) sec ^{2} ( u )}{2} d u

= 2 \cdot \int \frac{tan ^{4} ( u ) sec ^{2} ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int tan^{4} (u) sec^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \frac{1}{2} \cdot \int (1 + tan^{2} (u)) tan^{4} (u) d u

Apply u-substitution

: \int v^{4} d v

= 2 \cdot \frac{1}{2} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= 2 \cdot \frac{1}{2} \cdot \frac{v ^{5}}{5}

Substitute back

= 2 \cdot \frac{1}{2} \cdot \frac{tan ^{5} ( x ^{2} )}{5}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{tan ^{5} ( x ^{2} )}{5} : \frac{1}{5} tan^{5} (x^{2})

= \frac{1}{5} tan^{5} (x^{2})

Add a constant to the solution

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

\int 8 sin^{3} (θ) cos (θ) d θ

\int 3 x^{3} + 4 x^{2} - \frac{7}{9} d x

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

\int \frac{1}{4 y ^{2} + 4 y + 5} d y

\int 8 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

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