What is the integral of (sin(x))/(cos^2(x))x ?

The integral of (sin(x))/(cos^2(x))x is xsin(x)tan(x)+xcos(x)-ln|tan(x)+sec(x)|+C

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integral of (sin(x))/(cos^2(x))x

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

x sin (x) tan (x) + x cos (x) - ln ∣ tan (x) + sec (x) ∣ + C

Solution steps

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

Rewrite using trig identities

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

Apply Integration By Parts

: x sin (x) tan (x) - \int (sin (x) + x cos (x)) tan (x) d x

= x sin (x) tan (x) - \int (sin (x) + x cos (x)) tan (x) d x

\int (sin (x) + x cos (x)) tan (x) d x = - x cos (x) + ln ∣ tan (x) + sec (x) ∣

= x sin (x) tan (x) - (- x cos (x) + ln ∣ tan (x) + sec (x) ∣)

Simplify

= x sin (x) tan (x) + x cos (x) - ln ∣ tan (x) + sec (x) ∣

Add a constant to the solution

= x sin (x) tan (x) + x cos (x) - ln ∣ tan (x) + sec (x) ∣ + C

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

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

\int \frac{8}{x ^{2} - 2 x + 3} d x

\int \frac{x ^{4}}{4} \frac{1}{x} d x

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

What is the integral of (sin(x))/(cos^2(x))x ?
The integral of (sin(x))/(cos^2(x))x is xsin(x)tan(x)+xcos(x)-ln|tan(x)+sec(x)|+C