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Popular >

integral of (-8t^5)/(sqrt(t^2+2))

Solution

\int \frac{- 8 t ^{5}}{t ^{2} + 2} d t

Solution

- 322 \frac{1}{2} (2 + t^{2}) + \frac{32}{3} (2 + t^{2})^{\frac{3}{2}} - \frac{8}{5} (t^{2} + 2)^{\frac{5}{2}} + C

Solution steps

\int \frac{- 8 t ^{5}}{t ^{2} + 2} d t

Take the constant out:

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

= - 8 \cdot \int \frac{t ^{5}}{t ^{2} + 2} d t

Apply Trigonometric Substitution

: \int 42 tan^{5} (u) sec (u) d u

= - 8 \cdot \int 42 tan^{5} (u) sec (u) d u

Take the constant out:

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

= - 8 \cdot 42 \cdot \int tan^{5} (u) sec (u) d u

Rewrite using trig identities

= - 8 \cdot 42 \cdot \int (- 1 + sec^{2} (u))^{2} tan (u) sec (u) d u

Apply u-substitution

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

= - 8 \cdot 42 \cdot \int (- 1 + v^{2})^{2} d v

Expand

(- 1 + v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= - 8 \cdot 42 \cdot \int 1 - 2 v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 8 \cdot 42 (\int 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= - 8 \cdot 42 (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= - 8 \cdot 42 sec (arctan (\frac{1}{2} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} t ) )}{5}

Simplify

- 8 \cdot 42 sec (arctan (\frac{1}{2} t)) - \frac{2 sec ^{3} ( arctan ( \frac{1}{2} t ) )}{3} + \frac{sec ^{5} ( arctan ( \frac{1}{2} t ) )}{5} : - 322 \frac{1}{2} (2 + t^{2}) + \frac{32}{3} (2 + t^{2})^{\frac{3}{2}} - \frac{8}{5} (t^{2} + 2)^{\frac{5}{2}}

= - 322 \frac{1}{2} (2 + t^{2}) + \frac{32}{3} (2 + t^{2})^{\frac{3}{2}} - \frac{8}{5} (t^{2} + 2)^{\frac{5}{2}}

Add a constant to the solution

= - 322 \frac{1}{2} (2 + t^{2}) + \frac{32}{3} (2 + t^{2})^{\frac{3}{2}} - \frac{8}{5} (t^{2} + 2)^{\frac{5}{2}} + C

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