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

The integral of 1/(sqrt(1-sin^2(x))) is (ln|sin(x)+1|)/2-(ln|sin(x)-1|)/2+C

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

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

\frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} + C

Solution steps

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

Apply Trigonometric Substitution

= \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

Substitute back

u = sin (x)

= \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2}

Add a constant to the solution

= \frac{ln ∣ sin ( x ) + 1 ∣}{2} - \frac{ln ∣ sin ( x ) - 1 ∣}{2} + C

What is the integral of 1/(sqrt(1-sin^2(x))) ?
The integral of 1/(sqrt(1-sin^2(x))) is (ln|sin(x)+1|)/2-(ln|sin(x)-1|)/2+C