What is the integral of (arccos(x))/((1-x^2)^{3/2)} ?

The integral of (arccos(x))/((1-x^2)^{3/2)} is (xarccos(x))/(sqrt(1-x^2))-ln|sqrt(1-x^2)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (arccos(x))/((1-x^2)^{3/2)}

\int \frac{arccos ( x )}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x arccos ( x )}{1 - x ^{2}} - ln 1 - x^{2} + C

Solution steps

\int \frac{arccos ( x )}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int - \frac{u}{sin ^{2} ( u )} d u

= \int - \frac{u}{sin ^{2} ( u )} d u

Take the constant out:

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

= - \int \frac{u}{sin ^{2} ( u )} d u

Apply Integration By Parts

: - u cot (u) - \int - cot (u) d u

= - (- u cot (u) - \int - cot (u) d u)

\int - cot (u) d u = - ln ∣ sin (u) ∣

= - (- u cot (u) - (- ln ∣ sin (u) ∣))

Substitute back

u = arccos (x)

= - (- arccos (x) cot (arccos (x)) - (- ln ∣ sin (arccos (x)) ∣))

Simplify

- (- arccos (x) cot (arccos (x)) - (- ln ∣ sin (arccos (x)) ∣)) : \frac{x arccos ( x )}{1 - x ^{2}} - ln 1 - x^{2}

= \frac{x arccos ( x )}{1 - x ^{2}} - ln 1 - x^{2}

Add a constant to the solution

= \frac{x arccos ( x )}{1 - x ^{2}} - ln 1 - x^{2} + C

\int x^{2} - 3 x d x

\int \frac{1}{( x ^{2} + x ) ( x ^{2} + 9 )} d x

\int 4 (cos (x))^{2} d x

\int \frac{1}{sin ( x ) + 2 cos ( x ) + 3} d x

\int - 9 cos^{2} (x) d x

What is the integral of (arccos(x))/((1-x^2)^{3/2)} ?
The integral of (arccos(x))/((1-x^2)^{3/2)} is (xarccos(x))/(sqrt(1-x^2))-ln|sqrt(1-x^2)|+C