What is the integral of (3xsin^2(x))/(cos^4(x)) ?

The integral of (3xsin^2(x))/(cos^4(x)) is xtan^3(x)+ln|sec(x)|-1/2 sec^2(x)+C

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integral of (3xsin^2(x))/(cos^4(x))

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x sin ^{2} ( x )}{cos ^{4} ( x )} d x

Rewrite using trig identities

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

Expand

\frac{x ( 1 - cos ^{2} ( x ) )}{cos ^{4} ( x )} : \frac{x}{cos ^{4} ( x )} - \frac{x}{cos ^{2} ( x )}

= 3 \cdot \int \frac{x}{cos ^{4} ( x )} - \frac{x}{cos ^{2} ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int \frac{x}{cos ^{4} ( x )} d x - \int \frac{x}{cos ^{2} ( x )} d x)

\int \frac{x}{cos ^{4} ( x )} d x = x tan (x) + \frac{1}{3} x tan^{3} (x) + ln ∣ cos (x) ∣ + \frac{1}{3} ln ∣ sec (x) ∣ - \frac{1}{6} sec^{2} (x)

\int \frac{x}{cos ^{2} ( x )} d x = x tan (x) + ln ∣ cos (x) ∣

= 3 (x tan (x) + \frac{1}{3} x tan^{3} (x) + ln ∣ cos (x) ∣ + \frac{1}{3} ln ∣ sec (x) ∣ - \frac{1}{6} sec^{2} (x) - (x tan (x) + ln ∣ cos (x) ∣))

Simplify

3 (x tan (x) + \frac{1}{3} x tan^{3} (x) + ln ∣ cos (x) ∣ + \frac{1}{3} ln ∣ sec (x) ∣ - \frac{1}{6} sec^{2} (x) - (x tan (x) + ln ∣ cos (x) ∣)) : x tan^{3} (x) + ln ∣ sec (x) ∣ - \frac{1}{2} sec^{2} (x)

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

Add a constant to the solution

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

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

\int \frac{- x ^{2}}{4} d x

\int 4 - 4 x d x

\int \frac{2 x ^{2} - 6}{6 + 9 x - x ^{3}} d x

What is the integral of (3xsin^2(x))/(cos^4(x)) ?
The integral of (3xsin^2(x))/(cos^4(x)) is xtan^3(x)+ln|sec(x)|-1/2 sec^2(x)+C