What is the integral of 1/(sin(x)cos^3(x)) ?

The integral of 1/(sin(x)cos^3(x)) is (tan^2(x))/2+ln|tan(x)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sin(x)cos^3(x))

\int \frac{1}{sin ( x ) cos ^{3} ( x )} d x

\frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣ + C

Solution steps

\int \frac{1}{sin ( x ) cos ^{3} ( x )} d x

Apply Trigonometric Substitution

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

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

Simplify

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

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

Expand

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

= \int u + \frac{1}{u} 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 u d u + \int \frac{1}{u} d u

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

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + ln ∣ u ∣

Substitute back

u = tan (x)

= \frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣

Add a constant to the solution

= \frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣ + C

\int_{3}^{4} \frac{( 2 x - 3 )}{6} d x

1 - 9 x^{2} y^{^{'}} = x

\frac{d}{d x} (\frac{∣ x ∣}{2 - x ^{2}})

\frac{d y}{d t} + 5 y = 5

3 (1 + t^{2}) (\frac{d y}{d t}) = 2 t y (y^{3} - 1)

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