What is the integral of x^4tan^3(x^5) ?

The integral of x^4tan^3(x^5) is 1/10 (-ln(sec^2(x^5))+sec^2(x^5))+C

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integral of x^4tan^3(x^5)

\int x^{4} tan^{3} (x^{5}) d x

\frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

Solution steps

\int x^{4} tan^{3} (x^{5}) d x

Apply u-substitution

Take the constant out:

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

Apply u-substitution

: \int \frac{3 v ^{3}}{v ^{2} + 1} d v

= \frac{1}{15} \cdot \int \frac{3 v ^{3}}{v ^{2} + 1} d v

Take the constant out:

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

= \frac{1}{15} \cdot 3 \cdot \int \frac{v ^{3}}{v ^{2} + 1} d v

Apply u-substitution

: \int \frac{w - 1}{2 w} d w

= \frac{1}{15} \cdot 3 \cdot \int \frac{w - 1}{2 w} d w

Take the constant out:

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

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int \frac{w - 1}{w} d w

Expand

\frac{w - 1}{w} : 1 - \frac{1}{w}

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int 1 - \frac{1}{w} d w

Apply the Sum Rule:

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

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} (\int 1 d w - \int \frac{1}{w} d w)

\int 1 d w = w

\int \frac{1}{w} d w = ln ∣ w ∣

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} (w - ln ∣ w ∣)

Substitute back

Simplify

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5}))

Add a constant to the solution

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

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What is the integral of x^4tan^3(x^5) ?
The integral of x^4tan^3(x^5) is 1/10 (-ln(sec^2(x^5))+sec^2(x^5))+C