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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

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

Apply the Power Rule

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

= - \frac{1}{2 u ^{2}}

Substitute back

u = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

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