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

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

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integral of (tan^3(θ))/(cos^4(θ))

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

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

Solution steps

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

Rewrite using trig identities

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

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 (θ)

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

Add a constant to the solution

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

x y^{^{''}} + y^{^{'}} = 0

\frac{d}{d x} (π^{x})

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

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 4, y^{^{'}} (0) = 1

x \to - 9 - lim (x^{2} + \frac{3}{x + 9})

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