• May 14th 2010, 10:25 AM
alireza6485
Hi ,
Could you please help me with Integration : [sec sqrt(x)]^2 / 2 sqrt (x) ( 1+[tan sqrt(x)]^2]

First i simplified it :
sec sqrt(x)]^2 is equivalent to 1+[tan sqrt(x)]^2] I cancel them out.
1/2 sqrt(x) would remain.
Then the integration is equal to :
1/2 Integral x^ -1/2
( x^1/2)
I would really appreciate it if you show me the correct solution.
• May 14th 2010, 11:27 AM
Soroban
Hello, alireza64851

Quote:

$\int \frac{\sec^2\!\sqrt{x} }{2\sqrt{x}(1+\tan^2\!\sqrt{x})}\,dx$

First i simplified it :

. . $\sec^2\!\sqrt{x} \:=\:1+\tan^2\!\sqrt{x}$ . . . I cancelled them out.

Then: . $\frac{1}{2\sqrt{x}}$ would remain.

Then the integration is equal to: . $\tfrac{1}{2}\int x^{-\frac{1}{2}}dx$

The final answer is: . $x^{\frac{1}{2}} + C$