What is the integral of 31sec^{-2}(x)tan^3(x) ?

The integral of 31sec^{-2}(x)tan^3(x) is 31(1/2 cos^2(x)+ln|sec(x)|)+C

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integral of 31sec^{-2}(x)tan^3(x)

\int 31 sec^{- 2} (x) tan^{3} (x) d x

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

Solution steps

\int 31 sec^{- 2} (x) tan^{3} (x) d x

Take the constant out:

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

= 31 \cdot \int sec^{- 2} (x) tan^{3} (x) d x

Rewrite using trig identities

= 31 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec^{- 2} (x) d x

Apply u-substitution

= 31 \cdot \int \frac{- 1 + u ^{2}}{u ^{3}} d u

Expand

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

= 31 \cdot \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

= 31 (- \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 ∣

= 31 (- (- \frac{1}{2 u ^{2}}) + ln ∣ u ∣)

Substitute back

u = sec (x)

= 31 (- (- \frac{1}{2 sec ^{2} ( x )}) + ln ∣ sec (x) ∣)

Simplify

31 (- (- \frac{1}{2 sec ^{2} ( x )}) + ln ∣ sec (x) ∣) : 31 (\frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣)

= 31 (\frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣)

Add a constant to the solution

= 31 (\frac{1}{2} cos^{2} (x) + ln ∣ sec (x) ∣) + C

What is the integral of 31sec^{-2}(x)tan^3(x) ?
The integral of 31sec^{-2}(x)tan^3(x) is 31(1/2 cos^2(x)+ln|sec(x)|)+C