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

The integral of ((sin(x))/(1+tan(x)))^2 is -1/(4(1+tan(x)))-1/4 cos^2(x)-1/4 arctan(tan(x))+C

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integral of ((sin(x))/(1+tan(x)))^2

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

- \frac{1}{4 ( 1 + tan ( x ) )} - \frac{1}{4} cos^{2} (x) - \frac{1}{4} arctan (tan (x)) + C

Solution steps

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

Apply Trigonometric Substitution

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

Take the partial fraction of

\frac{u ^{2}}{( 1 + u ) ^{2} ( 1 + u ^{2} ) ^{2}} : \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u ^{2} + 1 )} + \frac{u}{2 ( u ^{2} + 1 ) ^{2}}

= \int \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u ^{2} + 1 )} + \frac{u}{2 ( u ^{2} + 1 ) ^{2}} d u

Apply the Sum Rule:

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

= \int \frac{1}{4 ( u + 1 ) ^{2}} d u - \int \frac{1}{4 ( u ^{2} + 1 )} d u + \int \frac{u}{2 ( u ^{2} + 1 ) ^{2}} d u

\int \frac{1}{4 ( u + 1 ) ^{2}} d u = - \frac{1}{4 ( u + 1 )}

\int \frac{1}{4 ( u ^{2} + 1 )} d u = \frac{1}{4} arctan (u)

\int \frac{u}{2 ( u ^{2} + 1 ) ^{2}} d u = - \frac{1}{4 ( u ^{2} + 1 )}

= - \frac{1}{4 ( u + 1 )} - \frac{1}{4} arctan (u) - \frac{1}{4 ( u ^{2} + 1 )}

Substitute back

u = tan (x)

= - \frac{1}{4 ( tan ( x ) + 1 )} - \frac{1}{4} arctan (tan (x)) - \frac{1}{4 ( tan ^{2} ( x ) + 1 )}

Simplify

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

= - \frac{1}{4 ( 1 + tan ( x ) )} - \frac{1}{4} cos^{2} (x) - \frac{1}{4} arctan (tan (x))

Add a constant to the solution

= - \frac{1}{4 ( 1 + tan ( x ) )} - \frac{1}{4} cos^{2} (x) - \frac{1}{4} arctan (tan (x)) + C

What is the integral of ((sin(x))/(1+tan(x)))^2 ?
The integral of ((sin(x))/(1+tan(x)))^2 is -1/(4(1+tan(x)))-1/4 cos^2(x)-1/4 arctan(tan(x))+C