What is the integral of 1/(sin(x)cos^5(x)) ?

The integral of 1/(sin(x)cos^5(x)) is (tan^4(x))/4+tan^2(x)+ln|tan(x)|+C

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integral of 1/(sin(x)cos^5(x))

\int \frac{1}{sin ( x ) cos ^{5} ( x )} d x

\frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣ + C

Solution steps

\int \frac{1}{sin ( x ) cos ^{5} ( x )} d x

Apply Trigonometric Substitution

: \int \frac{( u ^{2} + 1 ) ^{\frac{3}{2}} u ^{2} + 1}{u} d u

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

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

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

(u^{2} + 1) u^{2} + 1 u^{2} + 1 = (u^{2} + 1)^{2}

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

Expand

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

= \int u^{3} + 2 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^{3} d u + \int 2 u d u + \int \frac{1}{u} d u

\int u^{3} d u = \frac{u ^{4}}{4}

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

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{4}}{4} + u^{2} + ln ∣ u ∣

Substitute back

u = tan (x)

= \frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣

Add a constant to the solution

= \frac{tan ^{4} ( x )}{4} + tan^{2} (x) + ln ∣ tan (x) ∣ + C

\frac{d}{d x} (x (x - 7)^{2})

ln (7 - 3 x), 2

\frac{d}{d x} (\frac{7}{4 - 3 x})

\frac{d}{d x} (cos^{2} (a - x))

f (x) = \frac{x + x ^{3}}{x}

What is the integral of 1/(sin(x)cos^5(x)) ?
The integral of 1/(sin(x)cos^5(x)) is (tan^4(x))/4+tan^2(x)+ln|tan(x)|+C