What is the integral of 6tan^2(x)sec^4(x) ?

The integral of 6tan^2(x)sec^4(x) is 6((tan^3(x))/3+(tan^5(x))/5)+C

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integral of 6tan^2(x)sec^4(x)

\int 6 tan^{2} (x) sec^{4} (x) d x

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

Solution steps

\int 6 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

= 6 \cdot \int tan^{2} (x) sec^{4} (x) d x

Rewrite using trig identities

= 6 \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

= 6 \cdot \int u^{2} (1 + u^{2}) d u

Expand

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

= 6 \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

= 6 (\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}

= 6 (\frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (x)

= 6 (\frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5})

Add a constant to the solution

= 6 (\frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) + C

\frac{1}{( x ^{2} + x x )}

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What is the integral of 6tan^2(x)sec^4(x) ?
The integral of 6tan^2(x)sec^4(x) is 6((tan^3(x))/3+(tan^5(x))/5)+C