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

The integral of (2sin(x)cos(x))/(1+tan(x)) is

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{sin ( x ) cos ( x )}{1 + tan ( x )} d x

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

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

Substitute back

u = tan (x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + y^{^{'}} = 1

\int_{0}^{1} \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

(0, 2.5), (0.833, 0)

f (x) = 2 x^{3} - 2 x, at x = - 1

\frac{\partial}{\partial y} (sin (x + 2 y))

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