What is the integral of 1/((sin(x))^2(cos(x))^4) ?

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

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integral of 1/((sin(x))^2(cos(x))^4)

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

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

Solution steps

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

Simplify

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

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 = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{4}}{d x ^{4}} (ln (x))

15 x^{\frac{1}{4}}

n = 0 \sum \infty (\frac{2}{3})^{n - 1}

y^{^{'}} = \frac{x y ^{3}}{25}, y (0) = 10

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