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

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

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integral of sec^2(x)csc^4(x)

\int sec^{2} (x) csc^{4} (x) d x

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

Solution steps

\int sec^{2} (x) csc^{4} (x) d x

Rewrite using trig identities

= \int \frac{1}{sin ^{4} ( x ) cos ^{2} ( x )} d x

Apply Trigonometric Substitution

: \int - \frac{( 1 + u ^{2} ) ^{2}}{u ^{2}} d u

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

Take the constant out:

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

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

Expand

\frac{( 1 + u ^{2} ) ^{2}}{u ^{2}} : \frac{1}{u ^{2}} + 2 + u^{2}

= - \int \frac{1}{u ^{2}} + 2 + u^{2} d u

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{1}{u ^{2}} d u + \int 2 d u + \int u^{2} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int 2 d u = 2 u

\int u^{2} d u = \frac{u ^{3}}{3}

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

Substitute back

u = cot (x)

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

Simplify

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

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

Add a constant to the solution

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

y = 2 x - \frac{1}{2}

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\int x^{2} (5 x^{3} - 4)^{4} d x

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