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

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

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integral from 0 to 1/2 of ln(1-x^2)

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

\frac{1}{2} ln (\frac{3}{4}) + ln (3) - 1

Decimal

- 0.04522 \dots

Solution steps

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

Apply Integration By Parts

: [x ln (1 - x^{2}) - \int - \frac{2 x ^{2}}{1 - x ^{2}} d x]_{0}^{\frac{1}{2}}

= [x ln (1 - x^{2}) - \int - \frac{2 x ^{2}}{1 - x ^{2}} d x]_{0}^{\frac{1}{2}}

\int - \frac{2 x ^{2}}{1 - x ^{2}} d x = - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x

= [x ln (1 - x^{2}) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x)]_{0}^{\frac{1}{2}}

Simplify

= [x ln (1 - x^{2}) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{1}{2} ln (\frac{3}{4}) + ln (2) + ln (\frac{3}{2}) - 1

= \frac{1}{2} ln (\frac{3}{4}) + ln (2) + ln (\frac{3}{2}) - 1

Simplify

= \frac{1}{2} ln (\frac{3}{4}) + ln (3) - 1

\int_{0}^{\frac{π}{2}} sin^{4} (θ) d θ

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

\int_{1}^{2} (x^{3} + 3 x^{2} - 4 x) d x

\int_{0}^{4} 2 - x d x

\int_{0}^{0.5} 1 d x

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