What is the integral of 3tan^2(x)sec^4(x) ?

The integral of 3tan^2(x)sec^4(x) is tan^3(x)+3/5 tan^5(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3tan^2(x)sec^4(x)

\int 3 tan^{2} (x) sec^{4} (x) d x

tan^{3} (x) + \frac{3}{5} tan^{5} (x) + C

Solution steps

\int 3 tan^{2} (x) sec^{4} (x) d x

Take the constant out:

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

= 3 \cdot \int tan^{2} (x) sec^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{2} (1 + u^{2}) : u^{2} + u^{4}

= 3 \cdot \int u^{2} + u^{4} 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

= 3 (\int u^{2} d u + \int u^{4} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= 3 (\frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (x)

= 3 (\frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5})

Simplify

3 (\frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) : tan^{3} (x) + \frac{3}{5} tan^{5} (x)

= tan^{3} (x) + \frac{3}{5} tan^{5} (x)

Add a constant to the solution

= tan^{3} (x) + \frac{3}{5} tan^{5} (x) + C

\int \frac{y}{( y - 1 ) ^{4}} d y

\int \frac{x}{x ^{2} + 7 x + 13} d x

\int (4 x^{2} - 1)^{3} d x

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

\int \frac{1}{a - b x ^{2}} d x

What is the integral of 3tan^2(x)sec^4(x) ?
The integral of 3tan^2(x)sec^4(x) is tan^3(x)+3/5 tan^5(x)+C