1. ## easy integrals

1.
$\displaystyle \int \sec^2 x \csc^2 x dx$
2.
$\displaystyle \int^{\Pi/2}_{0} \log|\tan x| dx$

2. ## Re: easy integrals

1.

$\displaystyle \sec^2{x}\csc^2{x}=(\frac{1}{\sin{x}\cos{x}})^2=( \frac{2}{2\sin{x}\cos{x}})^2=( \frac{2}{\sin{2x}})^2=4\csc^2{(2x)}$

$\displaystyle \int \sec^2{x}\csc^2{x} dx = \int 4\csc^2{(2x)} dx$

Substitute 2x=u, and use identity $\displaystyle \int \csc^2{x} dx= -\cot{x}+C$

2. $\displaystyle \int_{0}^{\pi/2} \ln|\tan{x}|dx=0$ .

Reason:

Let ,$\displaystyle u=\ln|\tan{x}| \Leftrightarrow \frac{du}{dx}=\frac{\sec^2{x}}{\tan{x}} \Leftrightarrow dx=\frac{\tan{x}}{\sec^2{x}}du$

Also, when $\displaystyle x=0, u=\infty$ and $\displaystyle x=\pi/2,u=\infty$

Integral becomes $\displaystyle \int_{0}^{\pi/2} \ln|\tan{x}|dx=\int_{\infty}^{\infty}\frac{u\tan{x }}{\sec^2{x}}du=?$

3. ## Re: easy integrals

Originally Posted by sbhatnagar
2. $\displaystyle \int_{0}^{\pi/2} \ln|\tan{x}|dx=0$ .

Reason:

Let ,$\displaystyle u=\ln|\tan{x}| \Leftrightarrow \frac{du}{dx}=\frac{\sec^2{x}}{\tan{x}} \Leftrightarrow dx=\frac{\tan{x}}{\sec^2{x}}du$

Also, when $\displaystyle x=0, u=\infty$ and $\displaystyle x=\pi/2,u=\infty$

Integral becomes $\displaystyle \int_{0}^{\pi/2} \ln|\tan{x}|dx=\int_{\infty}^{\infty}\frac{u\tan{x }}{\sec^2{x}}du=?$
You need to finish the substitution to get x out of your integral.

If u = ln(tan(x)), then $\displaystyle \tan(x)=e^u\,.$

Also, $\displaystyle \sec^2(x)=1+\tan^2(x)=1+e^{2u}\,.$