Simple Integration problem

Dec 2008
509
2
Hi

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

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

P.S
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
  • Like
Reactions: TheCoffeeMachine
Dec 2008
509
2
Can you show me how you have done it, because i still cannot get the right answer.
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Dec 2008
509
2
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.
 
Jul 2007
894
298
New Orleans
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

\(\displaystyle u = \sin{x}\)

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