What is the integral of xsec^4(x^2+1)tan^7(x^2+1) ?

The integral of xsec^4(x^2+1)tan^7(x^2+1) is 1/2 ((tan^8(x^2+1))/8+(tan^{10}(x^2+1))/(10))+C

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integral of xsec^4(x^2+1)tan^7(x^2+1)

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

\frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10}) + C

Solution steps

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

Apply u-substitution

: \int \frac{tan ^{7} ( u ) sec ^{4} ( u )}{2} d u

= \int \frac{tan ^{7} ( u ) sec ^{4} ( 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^{7} (u) sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{7} (u) d u

Apply u-substitution

: \int v^{7} (1 + v^{2}) d v

= \frac{1}{2} \cdot \int v^{7} (1 + v^{2}) d v

Expand

v^{7} (1 + v^{2}) : v^{7} + v^{9}

= \frac{1}{2} \cdot \int v^{7} + v^{9} 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

= \frac{1}{2} (\int v^{7} d v + \int v^{9} d v)

\int v^{7} d v = \frac{v ^{8}}{8}

\int v^{9} d v = \frac{v ^{10}}{10}

= \frac{1}{2} (\frac{v ^{8}}{8} + \frac{v ^{10}}{10})

Substitute back

= \frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10})

Add a constant to the solution

= \frac{1}{2} (\frac{tan ^{8} ( x ^{2} + 1 )}{8} + \frac{tan ^{10} ( x ^{2} + 1 )}{10}) + C

x \to 3 lim (\frac{5}{x ( x ^{3} - 27 )})

\int \frac{8 x}{x ^{3} + 1} d x

x sin (π x)

\int cos (3 x) e^{2 s i n (3 x)} d x

\frac{d y}{d x} = x x - 30

What is the integral of xsec^4(x^2+1)tan^7(x^2+1) ?
The integral of xsec^4(x^2+1)tan^7(x^2+1) is 1/2 ((tan^8(x^2+1))/8+(tan^{10}(x^2+1))/(10))+C