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

The integral of tan^2(x)sin(x) is cos(x)+sec(x)+C

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integral of tan^2(x)sin(x)

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

cos (x) + sec (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{2}} d u

= \int - \frac{1 - u ^{2}}{u ^{2}} d u

Expand

- \frac{1 - u ^{2}}{u ^{2}} : - \frac{1}{u ^{2}} + 1

= \int - \frac{1}{u ^{2}} + 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

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

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int 1 d u = u

= - (- \frac{1}{u}) + u

Substitute back

u = cos (x)

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

Simplify

- (- \frac{1}{cos ( x )}) + cos (x) : cos (x) + sec (x)

= cos (x) + sec (x)

Add a constant to the solution

= cos (x) + sec (x) + C

f (x) = 7 x^{4}

\frac{\partial}{\partial x} (\frac{4 x y}{x ^{2} + y ^{2}})

x y^{^{'}} = y + (x + 1)^{2}

\int (x^{5} + x^{3})^{11} (5 x^{4} + 3 x^{2}) d x

\frac{( s - 1 )}{s ^{2} ( s ^{2} + 2 )}

What is the integral of tan^2(x)sin(x) ?
The integral of tan^2(x)sin(x) is cos(x)+sec(x)+C