What is the integral of (x^3csc^2(x^4))/(cot(x^4)) ?

The integral of (x^3csc^2(x^4))/(cot(x^4)) is -1/4 ln|cos(x^4)|+1/4 ln|sin(x^4)|+C

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integral of (x^3csc^2(x^4))/(cot(x^4))

\int \frac{x ^{3} csc ^{2} ( x ^{4} )}{cot ( x ^{4} )} d x

- \frac{1}{4} ln cos (x^{4}) + \frac{1}{4} ln sin (x^{4}) + C

Solution steps

\int \frac{x ^{3} csc ^{2} ( x ^{4} )}{cot ( x ^{4} )} d x

Rewrite using trig identities

= \int \frac{x ^{3} ( 1 + cot ^{2} ( x ^{4} ) )}{cot ( x ^{4} )} d x

Expand

\frac{x ^{3} ( 1 + cot ^{2} ( x ^{4} ) )}{cot ( x ^{4} )} : \frac{x ^{3}}{cot ( x ^{4} )} + x^{3} cot (x^{4})

= \int \frac{x ^{3}}{cot ( x ^{4} )} + x^{3} cot (x^{4}) d x

Apply the Sum Rule:

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

= \int \frac{x ^{3}}{cot ( x ^{4} )} d x + \int x^{3} cot (x^{4}) d x

\int \frac{x ^{3}}{cot ( x ^{4} )} d x = - \frac{1}{4} ln cos (x^{4})

\int x^{3} cot (x^{4}) d x = \frac{1}{4} ln sin (x^{4})

= - \frac{1}{4} ln cos (x^{4}) + \frac{1}{4} ln sin (x^{4})

Add a constant to the solution

= - \frac{1}{4} ln cos (x^{4}) + \frac{1}{4} ln sin (x^{4}) + C

What is the integral of (x^3csc^2(x^4))/(cot(x^4)) ?
The integral of (x^3csc^2(x^4))/(cot(x^4)) is -1/4 ln|cos(x^4)|+1/4 ln|sin(x^4)|+C