Hi
Can someone tell me how i would integrate the following question:
1) \(\displaystyle \int tan^2(x)\)
P.S
Can someone tell me how i would integrate the following question:
1) \(\displaystyle \int tan^2(x)\)
P.S
\(\displaystyle \int tan^2(x)dx\) = \(\displaystyle \int (sec^2(x) - 1)dx\)Hi
Can someone tell me how i would integrate the following question:
1) \(\displaystyle \int tan^2(x)\)
P.S
\(\displaystyle \int{\tan^2(x)}\;{dx} = \int{\left\{\sec^2(x)-1\right\}}\;{dx}\)1) \(\displaystyle \int tan^2(x)\)
Show your calculation.Can you show me how you have done it, because i still cannot get the right answer.
you picked the wrong u and vOk 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.
how can i find what v is equal to?you picked the wrong u and v
\(\displaystyle u = \sin{x}\)
\(\displaystyle dv = \frac{\sin{x}}{\cos^2{x}}dx\)
Similar Math Discussions | Math Forum | Date |
---|---|---|
simple integration problem | Calculus | |
SOLVED Simple integration problem | Calculus | |
Simple integration problem | Calculus | |
[SOLVED] Simple integration problem | Calculus |