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

The integral of 1/(1-sin^2(x))in(x) is dsec(x)+C

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integral of 1/(1-sin^2(x))in(x)

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

d sec (x) + C

Solution steps

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

Take the constant out:

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

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

Simplify

\frac{1}{1 - sin ^{2} ( x )} sin (x) : sec (x) tan (x)

= d \cdot \int sec (x) tan (x) d x

Apply u-substitution

= d \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= d \cdot 1 \cdot u

Substitute back

u = sec (x)

= d \cdot 1 \cdot sec (x)

Simplify

= d sec (x)

Add a constant to the solution

= d sec (x) + C

What is the integral of 1/(1-sin^2(x))in(x) ?
The integral of 1/(1-sin^2(x))in(x) is dsec(x)+C