What is the integral of xtan^4(x^2) ?

The integral of xtan^4(x^2) is 1/2 (1/3 tan^3(x^2)+x^2-tan(x^2))+C

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integral of xtan^4(x^2)

\int x tan^{4} (x^{2}) d x

\frac{1}{2} (\frac{1}{3} tan^{3} (x^{2}) + x^{2} - tan (x^{2})) + C

Solution steps

\int x tan^{4} (x^{2}) d x

Apply u-substitution

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

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

Apply Integral Reduction

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

\int tan^{2} (u) d u = - u + tan (u)

= \frac{1}{2} (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = x^{2}

= \frac{1}{2} (\frac{tan ^{3} ( x ^{2} )}{3} - (- x^{2} + tan (x^{2})))

Simplify

\frac{1}{2} (\frac{tan ^{3} ( x ^{2} )}{3} - (- x^{2} + tan (x^{2}))) : \frac{1}{2} (\frac{1}{3} tan^{3} (x^{2}) + x^{2} - tan (x^{2}))

= \frac{1}{2} (\frac{1}{3} tan^{3} (x^{2}) + x^{2} - tan (x^{2}))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} tan^{3} (x^{2}) + x^{2} - tan (x^{2})) + C

\int \frac{6}{x ^{4}} d x

x \to - 2 - lim (\frac{- 6 x}{x + 2})

\int_{0}^{7} \frac{1}{49 - t ^{2}} d t

\int_{0}^{2} \frac{2 x}{( 1 + x ^{2} ) ^{2}} d x

\int 3 x^{3} 4 x^{4} d x

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