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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

\int_{0}^{3} x^{2} - 3 x d x

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

\int_{1}^{8} 9 \frac{ln ( x )}{x ^{2}} d x

x^{x^{2}}, 1

\frac{1}{x x ^{2} - 1}

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