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

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

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

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

\frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣ + C

Solution steps

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

Rewrite using trig identities

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

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

Simplify

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

= \frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣

Add a constant to the solution

= \frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣ + C

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\int \frac{5 x}{( 2 - x ^{2} ) ^{3}} d x

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