What is the integral of tan^{-3}(x)sec^4(x) ?

The integral of tan^{-3}(x)sec^4(x) is -1/2 cot^2(x)+ln|tan(x)|+C

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integral of tan^{-3}(x)sec^4(x)

\int tan^{- 3} (x) sec^{4} (x) d x

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

Solution steps

\int tan^{- 3} (x) sec^{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 \frac{1 + u ^{2}}{u ^{3}} d u

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

Expand

\frac{1 + u ^{2}}{u ^{3}} : \frac{1}{u ^{3}} + \frac{1}{u}

= \int \frac{1}{u ^{3}} + \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 \frac{1}{u ^{3}} d u + \int \frac{1}{u} d u

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{1}{u} d u = ln ∣ u ∣

= - \frac{1}{2 u ^{2}} + ln ∣ u ∣

Substitute back

u = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of tan^{-3}(x)sec^4(x) ?
The integral of tan^{-3}(x)sec^4(x) is -1/2 cot^2(x)+ln|tan(x)|+C