What is the integral of (15)/(1-sin^2(x)) ?

The integral of (15)/(1-sin^2(x)) is 15tan(x)+C

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integral of (15)/(1-sin^2(x))

\int \frac{15}{1 - sin ^{2} ( x )} d x

15 tan (x) + C

Solution steps

\int \frac{15}{1 - sin ^{2} ( x )} d x

Take the constant out:

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

= 15 \cdot \int \frac{1}{1 - sin ^{2} ( x )} d x

Rewrite using trig identities

= 15 \cdot \int sec^{2} (x) d x

Use the common integral:

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

= 15 tan (x)

Add a constant to the solution

= 15 tan (x) + C

\frac{2 ^{x}}{3 x - 1}

\int_{0}^{5} \frac{x}{1 + 3 x} d x

\int \frac{1}{x x ^{2} - 1} d x

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

(lo g_{2} (x))^{2}

What is the integral of (15)/(1-sin^2(x)) ?
The integral of (15)/(1-sin^2(x)) is 15tan(x)+C