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

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

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

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

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

Decimal

1.14779 \dots

Solution steps

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

Rewrite using trig identities

= \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} 3 x^{3} - 9 x^{2} + 6 x d x

\int_{0}^{1} x (1 - x^{2})^{9} d x

\int_{0}^{3} 400 - 20 t d t

\int_{1.4}^{1.9} \frac{1}{2} x d x

\int_{4}^{8} 6 x d x

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