What is the derivative of the trigonometric integral?

[karush]

$$\int_{0 }^{\pi/2 } x\sec^2 \left({{x}^{2}}\right)\tan\left({{x}^{2}}\right)\,dx$$
Not sure where to start on this...
Except that

$$\d{}{x}\sec^{2 } \left({x^2 }\right)=\frac{4x\sin\left({{x}^{2}}\right)}{\cos^{3}\left({{x}^{2}}\right)}$$

 

[MarkFL]

I would begin with:

$$u=x^2\,\therefore\,du=2x\,dx$$

and continue with:

$$v=\tan(u)\,\therefore\,dv=\sec^2(u)\,du$$

(Don't forget to change the limits in accordance with the substitutions...) :)

edit: Beware of singularities (improper integral)!

 

[karush]

I did notice a VA in the interval, does mean that it's an improper integral

I didn't know how to find where the VA was using limits.

 

[MarkFL]

Let's make the first substitution I suggested, and we then get:

$$I=\frac{1}{2}\int_0^{\frac{\pi^2}{4}}\tan(u)\sec^2(u)\,du$$

Now, we know that both the tangent and secant functions are undefined when their arguments are $$\frac{\pi}{2}$$, and we know that since:

$$1<\frac{\pi}{2}$$

then:

$$\frac{\pi}{2}<\left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{4}$$

And so we must therefore write:

$$I=\frac{1}{2}\left(\lim_{t\to\frac{\pi}{2}^{-}}\left(\int_0^{t}\tan(u)\sec^2(u)\,du\right)+\lim_{t\to\frac{\pi}{2}^{+}}\left(\int_t^{\frac{\pi^2}{4}}\tan(u)\sec^2(u)\,du\right)\right)$$

What do you get when you now apply the second substitution I suggested?

 

[karush]

Just curious what happened to the x the original given?

 

[MarkFL]

Just curious what happened to the x the original given?

I used the substitution $$u=x^2$$ as given in my first post to rewrite the given definite integral in terms of $u$ instead of $x$. Having $x$ as one of the factors in the integrand made this possible. This way we do not have squared arguments for the trig. functions. :)

 

[Greg]

A different approach:

Starting with

$$\int x\sec^2x^2\tan x^2\,dx$$

make the substitution

$$u=x^2$$.

Then

$$\dfrac12\,du=x\,dx$$

so we have

$$\dfrac12\int\sec^2u\tan u\,du$$.

Make the substitution

$$w=\tan u$$.

Then

$$dw=\sec^2u\,du$$

so we have

$$\dfrac12\int w\,dw=\dfrac{w^2}{4}+C$$.

Back-substitute:

$$\int x\sec^2x^2\tan x^2\,dx=\dfrac{\tan^2x^2}{4}+C$$

For this expression, we have a singularity at $x=\sqrt{\dfrac{\pi}{2}}$

so we look at the left and right-hand limits around this value:

$\lim_{x\to\left(\sqrt{\dfrac{\pi}{2}}\right)^-}\dfrac{\tan^2x^2}{4}=+\infty$ and $\lim_{x\to\left(\sqrt{\dfrac{\pi}{2}}\right)^+}\dfrac{\tan^2x^2}{4}=-\infty$

and we conclude that the integral does not converge.

I believe this is a valid approach.

 

[karush]

Great help, I would have never gotten that.

the book examples were foggy. Friends no help.

So if you want serious help you come here.

☺☺☺

 

[MarkFL]

I get the same result as greg1313. Recall, I left it at:

$$I=\frac{1}{2}\left(\lim_{t\to\frac{\pi}{2}^{-}}\left(\int_0^{t}\tan(u)\sec^2(u)\,du\right)+\lim_{t\to\frac{\pi}{2}^{+}}\left(\int_t^{\frac{\pi^2}{4}}\tan(u)\sec^2(u)\,du\right)\right)$$

Making the second substitution, we obtain:

$$I=\frac{1}{2}\left(\lim_{t\to\frac{\pi}{2}^{-}}\left(\int_0^{\tan(t)}v\,dv\right)+\lim_{t\to\frac{\pi}{2}^{+}}\left(\int_{\tan(t)}^{\tan\left(\frac{\pi^2}{4}\right)}v\,dv\right)\right)$$

Applying the FTOC, there results:

$$I=\frac{1}{4}\left(\lim_{t\to\frac{\pi}{2}^{-}}\left(\left[v^2\right]_0^{\tan(t)}\right)+\lim_{t\to\frac{\pi}{2}^{+}}\left(\left[v^2\right]_{\tan(t)}^{\tan\left(\frac{\pi^2}{4}\right)}\right)\right)$$

$$I=\frac{1}{4}\left(\lim_{t\to\frac{\pi}{2}^{-}}\left(\tan^2(t)\right)+\lim_{t\to\frac{\pi}{2}^{+}}\left(\tan^2\left(\frac{\pi^2}{4}\right)-\tan^2(t)\right)\right)$$

Recall that whenever we separate an improper integral into smaller integrals, the improper integral is convergent iff the two parts of the integral are convergent. Thus, as tempting as it may be to "cancel" the resulting infinities, we must reckon that the given integral does not converge.

 

[suluclac]

To avoid confusion, I'll discuss only the indefinite integral.
Let v = sec u.
$$\int x\sec^2{x^2}\tan{x^2}\,dx=\frac{\sec^2{x^2}}{4}\,+\,C$$
Take the derivative for both this and the tangent squared and we'll arrive with the same integrand.