What is the integral of (1+xtan(x)-sin(x))sec^2(x) ?

The integral of (1+xtan(x)-sin(x))sec^2(x) is 1/2 xtan^2(x)+1/2 x+1/2 tan(x)-sec(x)+C

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integral of (1+xtan(x)-sin(x))sec^2(x)

integral

(1 + x tan (x) - sin (x)) sec^{2} (x)

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

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( x ) + x sin ( x ) - sin ( x ) cos ( x )}{cos ^{3} ( x )} d x

Expand

\frac{cos ( x ) + x sin ( x ) - sin ( x ) cos ( x )}{cos ^{3} ( x )} : \frac{1}{cos ^{2} ( x )} + \frac{x sin ( x )}{cos ^{3} ( x )} - \frac{sin ( x )}{cos ^{2} ( x )}

= \int \frac{1}{cos ^{2} ( x )} + \frac{x sin ( x )}{cos ^{3} ( x )} - \frac{sin ( 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

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

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

\int \frac{x sin ( x )}{cos ^{3} ( x )} d x = \frac{1}{2} x tan^{2} (x) - \frac{1}{2} (- x + tan (x))

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

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

Simplify

tan (x) + \frac{1}{2} x tan^{2} (x) - \frac{1}{2} (- x + tan (x)) - sec (x) : \frac{1}{2} x tan^{2} (x) + \frac{1}{2} x + \frac{1}{2} tan (x) - sec (x)

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

Add a constant to the solution

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

cos^{2} (x) - sin^{2} (x)

\frac{( 1 - x )}{( x + 2 )}

5 + 2 x

\frac{( 1 + sin ( x ) )}{cos ( x )}

\frac{( x - 1 )}{( x + 1 )}

What is the integral of (1+xtan(x)-sin(x))sec^2(x) ?
The integral of (1+xtan(x)-sin(x))sec^2(x) is 1/2 xtan^2(x)+1/2 x+1/2 tan(x)-sec(x)+C