What is the integral of (5x)/(3-3sin^2(1-x^2)) ?

The integral of (5x)/(3-3sin^2(1-x^2)) is -5/6 tan(-x^2+1)+C

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integral of (5x)/(3-3sin^2(1-x^2))

\int \frac{5 x}{3 - 3 sin ^{2} ( 1 - x ^{2} )} d x

- \frac{5}{6} tan (- x^{2} + 1) + C

Solution steps

\int \frac{5 x}{3 - 3 sin ^{2} ( 1 - x ^{2} )} d x

Take the constant out:

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

= 5 \cdot \int \frac{x}{3 - 3 sin ^{2} ( 1 - x ^{2} )} d x

Rewrite using trig identities

= 5 \cdot \int \frac{x}{3 cos ^{2} ( - x ^{2} + 1 )} d x

Take the constant out:

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

= 5 \cdot \frac{1}{3} \cdot \int \frac{x}{cos ^{2} ( - x ^{2} + 1 )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= 5 \cdot \frac{1}{3} \cdot \int x sec^{2} (- x^{2} + 1) d x

Apply u-substitution

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

= 5 \cdot \frac{1}{3} \cdot \int - \frac{sec ^{2} ( u )}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{3} (- \frac{1}{2} \cdot \int sec^{2} (u) d u)

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 5 \cdot \frac{1}{3} (- \frac{1}{2} tan (u))

Substitute back

u = - x^{2} + 1

= 5 \cdot \frac{1}{3} (- \frac{1}{2} tan (- x^{2} + 1))

Simplify

5 \cdot \frac{1}{3} (- \frac{1}{2} tan (- x^{2} + 1)) : - \frac{5}{6} tan (- x^{2} + 1)

= - \frac{5}{6} tan (- x^{2} + 1)

Add a constant to the solution

= - \frac{5}{6} tan (- x^{2} + 1) + C

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What is the integral of (5x)/(3-3sin^2(1-x^2)) ?
The integral of (5x)/(3-3sin^2(1-x^2)) is -5/6 tan(-x^2+1)+C