# Simple Integration problem

#### Paymemoney

Hi

Can someone tell me how i would integrate the following question:

1) $$\displaystyle \int tan^2(x)$$

P.S

#### sa-ri-ga-ma

Hi

Can someone tell me how i would integrate the following question:

1) $$\displaystyle \int tan^2(x)$$

P.S
$$\displaystyle \int tan^2(x)dx$$ = $$\displaystyle \int (sec^2(x) - 1)dx$$

#### TheCoffeeMachine

1) $$\displaystyle \int tan^2(x)$$
$$\displaystyle \int{\tan^2(x)}\;{dx} = \int{\left\{\sec^2(x)-1\right\}}\;{dx}$$

EDIT: Bit late.

#### chisigma

MHF Hall of Honor
Because is...

$$\displaystyle \tan^{2} x = \sin x \cdot \frac{\sin x}{\cos^{2} x}$$ (1)

... You can integrate by parts and obtain...

$$\displaystyle \int \tan^{2} x \cdot dx = \frac{\sin x}{\cos x} - \int dx = \tan x - x + c$$ (2)

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

TheCoffeeMachine

#### Paymemoney

Can you show me how you have done it, because i still cannot get the right answer.

#### sa-ri-ga-ma

Can you show me how you have done it, because i still cannot get the right answer.

Do you know $$\displaystyle \int\sec^2{x}dx$$ ?

and $$\displaystyle \int{dx}$$ ?

#### chisigma

MHF Hall of Honor
Integration by parts:

$$\displaystyle \int f(x)\cdot g(x)\cdot dx = f(x)\cdot G(x) - \int G(x)\cdot f^{'} (x)\cdot dx$$ (1)

... where $$\displaystyle G(*)$$ is a primitive of $$\displaystyle g(*)$$. In our case is $$\displaystyle f(x)= \sin x$$ and $$\displaystyle g(x)= \frac{\sin x}{\cos^{2} x}$$ ...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### Paymemoney

Ok this is what i have done:

$$\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$$

$$\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$$

$$\displaystyle dv = sin(x) v = -cos(x)$$

$$\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$$

$$\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$$

Someone tell me where my mistake is.

#### 11rdc11

Ok this is what i have done:

$$\displaystyle \int tan^2(x) = \int sinx * \frac{sin}{cos^2(x)}$$

$$\displaystyle u= \frac{sin(x)}{cos^2(x)} du = \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)}dx$$

$$\displaystyle dv = sin(x) v = -cos(x)$$

$$\displaystyle \frac{sin(x)}{cos^2(x)} * -cos(x) - \int \frac{cos^3(x) - 2cos(x)sin2(x)}{cos^4(x)} * -cos(x)$$

$$\displaystyle tan(x) + 2 \int cos^2(x)sin^2(x)$$

Someone tell me where my mistake is.
you picked the wrong u and v

$$\displaystyle u = \sin{x}$$

$$\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$$

#### Paymemoney

you picked the wrong u and v

$$\displaystyle u = \sin{x}$$

$$\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx$$
how can i find what v is equal to?