I on this indefinite integral

Oomair
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Homework Statement


ok I am given this problem

indef. int (1+tan^2*5x)dx i need to use the u subsitution method to find the answer but i cannot seem to find what to subsitute

the worksheet says the answer is " one-fifth*tan5x+C


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The Attempt at a Solution

 
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The using the Pythagorean Identity 1+ \tan^2 x = \sec^2 x, we can change your problem to:
\int \sec^2 (5x) dx. then let u= 5x. du= 5 dx, or dx = (1/5) du.

We can take constants out of the integral, so it becomes \frac{1}{5} \int \sec^2 u du. You should know that the derivative of tan x is sec^2 x, so the integral is \frac{1}{5} \tan u + C = \frac{1}{5} \tan (5x) + C
 
thanks for the help, i got the trig idenity, but the problem was that i was letting u=sec^25x
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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