What is the integral from 0 to pi/4 of 1/(cos^3(x)) ?

The integral from 0 to pi/4 of 1/(cos^3(x)) is (ln(1+sqrt(2))+sqrt(2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/4 of 1/(cos^3(x))

\int_{0}^{\frac{π}{4}} \frac{1}{cos ^{3} ( x )} d x

\frac{ln ( 1 + 2 ) + 2}{2}

Decimal

1.14779 \dots

Solution steps

\int_{0}^{\frac{π}{4}} \frac{1}{cos ^{3} ( x )} d x

\frac{1}{cos ^{3} ( x )} = sec^{3} (x) (1 - cos^{3} (x)) + 1

= \int_{0}^{\frac{π}{4}} sec^{3} (x) (1 - cos^{3} (x)) + 1 d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{1 - cos ^{3} ( x )}{cos ^{3} ( x )} + 1 d x

Apply Integration By Parts

: [x (sec^{3} (x) (1 - cos^{3} (x)) + 1) - \int 3 x sec^{4} (x) sin (x) d x]_{0}^{\frac{π}{4}}

= [x (sec^{3} (x) (1 - cos^{3} (x)) + 1) - \int 3 x sec^{4} (x) sin (x) d x]_{0}^{\frac{π}{4}}

\int 3 x sec^{4} (x) sin (x) d x = x sec^{3} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= [x (sec^{3} (x) (1 - cos^{3} (x)) + 1) - (x sec^{3} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)]_{0}^{\frac{π}{4}}

Simplify

[x (sec^{3} (x) (1 - cos^{3} (x)) + 1) - (x sec^{3} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)]_{0}^{\frac{π}{4}} : [\frac{1}{2} (- 2 x sec^{3} (x) cos^{3} (x) + 2 x + sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣)]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (- 2 x sec^{3} (x) cos^{3} (x) + 2 x + sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{ln ( 1 + 2 ) + 2}{2}

= \frac{ln ( 1 + 2 ) + 2}{2}

\int_{1}^{2} (x^{2} - 2 x + 3) d x

\int_{- 1}^{1} sin^{3} (\frac{x}{x ^{2} + 1}) d x

\int_{0}^{\infty} \frac{1}{2 x + 1} d x

\int_{0}^{1} 2 x^{2} - 2 x d x

\int_{0}^{0.5} x^{2} e^{- x^{2}} d x

What is the integral from 0 to pi/4 of 1/(cos^3(x)) ?
The integral from 0 to pi/4 of 1/(cos^3(x)) is (ln(1+sqrt(2))+sqrt(2))/2