What is the integral from 0 to pi/4 of x/(cos^2(x)) ?

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

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integral from 0 to pi/4 of x/(cos^2(x))

\int_{0}^{\frac{π}{4}} \frac{x}{cos ^{2} ( x )} d x

\frac{π}{4} - \frac{1}{2} ln (2)

Decimal

0.43882 \dots

Solution steps

\int_{0}^{\frac{π}{4}} \frac{x}{cos ^{2} ( x )} d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} \frac{2 x}{1 + cos ( 2 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{\frac{π}{4}} \frac{x}{1 + cos ( 2 x )} d x

Apply Integration By Parts

: [\frac{1}{2} (x tan (x) - 2 \cdot \int \frac{1}{2} tan (x) d x)]_{0}^{\frac{π}{4}}

= 2 [\frac{1}{2} (x tan (x) - 2 \cdot \int \frac{1}{2} tan (x) d x)]_{0}^{\frac{π}{4}}

\int \frac{1}{2} tan (x) d x = - \frac{1}{2} ln ∣ cos (x) ∣

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2} (\frac{π}{4} - \frac{1}{2} ln (2))

= 2 \cdot \frac{1}{2} (\frac{π}{4} - \frac{1}{2} ln (2))

Simplify

= \frac{π}{4} - \frac{1}{2} ln (2)

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