What is the integral of tan^{-3}(x)sec^2(x)sec^2(x) ?

The integral of tan^{-3}(x)sec^2(x)sec^2(x) is -1/2 cot^2(x)-ln|cot(x)|+C

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integral of tan^{-3}(x)sec^2(x)sec^2(x)

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

- \frac{1}{2} cot^{2} (x) - ln ∣ cot (x) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{sin ^{3} ( x ) cos ( x )} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Simplify

(u^{2} + 1)^{\frac{1}{2}} u^{2} + 1 : u^{2} + 1

= - \int \frac{u ^{2} + 1}{u} d u

Expand

\frac{u ^{2} + 1}{u} : u + \frac{1}{u}

= - \int u + \frac{1}{u} 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 d u + \int \frac{1}{u} d u)

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - (\frac{u ^{2}}{2} + ln ∣ u ∣)

Substitute back

u = cot (x)

= - (\frac{cot ^{2} ( x )}{2} + ln ∣ cot (x) ∣)

Simplify

- (\frac{cot ^{2} ( x )}{2} + ln ∣ cot (x) ∣) : - \frac{1}{2} cot^{2} (x) - ln ∣ cot (x) ∣

= - \frac{1}{2} cot^{2} (x) - ln ∣ cot (x) ∣

Add a constant to the solution

= - \frac{1}{2} cot^{2} (x) - ln ∣ cot (x) ∣ + C

x \to 0 lim (\frac{sin ( x ^{8} )}{x})

tan (e^{7 t}) + e^{t a n (7 t)}

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

y = e^{\frac{x}{3}}, at x = 0

\int cos^{2} (\frac{2 x}{3}) d x

What is the integral of tan^{-3}(x)sec^2(x)sec^2(x) ?
The integral of tan^{-3}(x)sec^2(x)sec^2(x) is -1/2 cot^2(x)-ln|cot(x)|+C