What is the integral of 4/((4-(2sin(θ))^2)^{3/2)} ?

The integral of 4/((4-(2sin(θ))^2)^{3/2)} is 1/2 (1/2 sec(θ)tan(θ)+1/2 ln|tan(θ)+sec(θ)|)+C

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integral of 4/((4-(2sin(θ))^2)^{3/2)}

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

\frac{1}{2} (\frac{1}{2} sec (θ) tan (θ) + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{( 4 - ( 2 sin ( θ ) ) ^{2} ) ^{\frac{3}{2}}} d θ

Simplify

\frac{1}{( 4 - ( 2 sin ( θ ) ) ^{2} ) ^{\frac{3}{2}}} : \frac{1}{8} sec^{3} (θ)

= 4 \cdot \int \frac{1}{8} sec^{3} (θ) d θ

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int sec^{3} (θ) d θ

Apply Integral Reduction

= 4 \cdot \frac{1}{8} (\frac{sec ^{2} ( θ ) sin ( θ )}{2} + \frac{1}{2} \cdot \int sec (θ) d θ)

\int sec (θ) d θ = ln ∣ tan (θ) + sec (θ) ∣

= 4 \cdot \frac{1}{8} (\frac{sec ^{2} ( θ ) sin ( θ )}{2} + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣)

Simplify

4 \cdot \frac{1}{8} (\frac{sec ^{2} ( θ ) sin ( θ )}{2} + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣) : \frac{1}{2} (\frac{1}{2} sec (θ) tan (θ) + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣)

= \frac{1}{2} (\frac{1}{2} sec (θ) tan (θ) + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣)

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sec (θ) tan (θ) + \frac{1}{2} ln ∣ tan (θ) + sec (θ) ∣) + C

\int \frac{( 5 x ^{3} - 6 x + 4 )}{x ^{3} - x ^{2}} d x

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\int \frac{1}{x 1 + 2 ln ( x )} d x

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

\int cos^{2} (s) d s

What is the integral of 4/((4-(2sin(θ))^2)^{3/2)} ?
The integral of 4/((4-(2sin(θ))^2)^{3/2)} is 1/2 (1/2 sec(θ)tan(θ)+1/2 ln|tan(θ)+sec(θ)|)+C