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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

= \int \frac{2 u}{1 + cos ( 2 u )} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{1 + cos ( 2 u )} d u

Apply Integration By Parts

: \frac{1}{2} u tan (u) - \int \frac{1}{2} tan (u) d u

= 2 (\frac{1}{2} u tan (u) - \int \frac{1}{2} tan (u) d u)

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

= 2 (\frac{1}{2} u tan (u) - (- \frac{1}{2} ln ∣ cos (u) ∣))

Substitute back

u = arcsin (x)

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

Simplify

2 (\frac{1}{2} arcsin (x) tan (arcsin (x)) - (- \frac{1}{2} ln ∣ cos (arcsin (x)) ∣)) : \frac{x arcsin ( x )}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + ln 1 - x^{2}

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

Add a constant to the solution

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

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