What is the integral of xtan^8(x^2)sec^2(x^2) ?

The integral of xtan^8(x^2)sec^2(x^2) is 1/18 tan^9(x^2)+C

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integral of xtan^8(x^2)sec^2(x^2)

\int x tan^{8} (x^{2}) sec^{2} (x^{2}) d x

\frac{1}{18} tan^{9} (x^{2}) + C

Solution steps

\int x tan^{8} (x^{2}) sec^{2} (x^{2}) d x

Apply u-substitution

: \int \frac{tan ^{8} ( u ) sec ^{2} ( u )}{2} d u

= \int \frac{tan ^{8} ( u ) sec ^{2} ( u )}{2} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{8} d v

= \frac{1}{2} \cdot \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= \frac{1}{2} \cdot \frac{v ^{9}}{9}

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{9} ( x ^{2} )}{9}

Simplify

\frac{1}{2} \cdot \frac{tan ^{9} ( x ^{2} )}{9} : \frac{1}{18} tan^{9} (x^{2})

= \frac{1}{18} tan^{9} (x^{2})

Add a constant to the solution

= \frac{1}{18} tan^{9} (x^{2}) + C

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What is the integral of xtan^8(x^2)sec^2(x^2) ?
The integral of xtan^8(x^2)sec^2(x^2) is 1/18 tan^9(x^2)+C