What is the integral of (tan^3(x))/(cos^4(x)) ?

The integral of (tan^3(x))/(cos^4(x)) is (tan^4(x))/4+(tan^6(x))/6+C

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integral of (tan^3(x))/(cos^4(x))

\int \frac{tan ^{3} ( x )}{cos ^{4} ( x )} d x

\frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6} + C

Solution steps

\int \frac{tan ^{3} ( x )}{cos ^{4} ( x )} d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

u^{3} (1 + u^{2}) : u^{3} + u^{5}

= \int u^{3} + u^{5} 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^{3} d u + \int u^{5} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = tan (x)

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6}

Add a constant to the solution

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{6} + C

\frac{\partial}{\partial x} (ln ((\frac{y}{x})^{2}))

\int \frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\frac{\partial}{\partial x} (41 - x^{2} - y^{2})

\frac{\partial}{\partial x} (sin (2 y))

x \to 1 lim (∣ x ∣)

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