What is the integral of 1/(3-4sin^2(x)) ?

The integral of 1/(3-4sin^2(x)) is 1/(2sqrt(3))(ln|(tan(x))/(sqrt(3))+1|-ln|(tan(x))/(sqrt(3))-1|)+C

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integral of 1/(3-4sin^2(x))

\int \frac{1}{3 - 4 sin ^{2} ( x )} d x

\frac{1}{2 3} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1) + C

Solution steps

\int \frac{1}{3 - 4 sin ^{2} ( x )} d x

Apply Trigonometric Substitution

: \int \frac{1}{3 - u ^{2}} d u

= \int \frac{1}{3 - u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{3 ( 1 - v ^{2} )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{1 - v ^{2}} d v

Use the common integral:

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

= \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \frac{ln \frac{t a n ( x )}{3} + 1}{2} - \frac{ln \frac{t a n ( x )}{3} - 1}{2}

Simplify

\frac{1}{3} \frac{ln \frac{t a n ( x )}{3} + 1}{2} - \frac{ln \frac{t a n ( x )}{3} - 1}{2} : \frac{1}{2 3} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1)

= \frac{1}{2 3} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1)

Add a constant to the solution

= \frac{1}{2 3} (ln \frac{tan ( x )}{3} + 1 - ln \frac{tan ( x )}{3} - 1) + C

f (x) = x^{2020}

\int \frac{2 x ^{2}}{36 - x ^{2}} d x

\int 2 sec (x) tan (x) - 5 sec^{2} (x) d x

\frac{d}{d x} ((x^{3} + 5 x)^{2} (5 x - 3)^{4})

f (x) = tan (7 x)

What is the integral of 1/(3-4sin^2(x)) ?
The integral of 1/(3-4sin^2(x)) is 1/(2sqrt(3))(ln|(tan(x))/(sqrt(3))+1|-ln|(tan(x))/(sqrt(3))-1|)+C