Thread: Calculus Trig Integration/substitution, by parts, improper integrals?

1. Calculus Trig Integration/substitution, by parts, improper integrals?

) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx

2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

3) 0 to pi ∫ e^(cosx)sin(2x) dx

4) 4 to 9∫ ln(x)/sqrt(x) dx

5) Derive the basic Integral

∫csc(x)dx=ln[cot(x)-csc(x)]+C

Help on any of these would be appreciated.

2. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Just in case a picture helps with 5, for a start... integral of cosec

3. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

I'm not sure how to interpret that haha, could you explain a little?

4. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

That particular pic has a step-by-step - click the step-forward button.

Here's some 'lazy integration by parts'...

... where (key in spoiler) ...

Spoiler:

... is the product rule, where straight continuous lines are differentiating downwards with respect to x.

The general drift is...

... i.e. lazy integration by parts, doing without u and v.

Here's some more...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of the product rule. (See previous.)

Full size

In all cases, start from bottom left, work anti-clockwise (kinda).

Enjoy!!

PS, no.2 blows up half way along from zero to pi/4. Was it perhaps meant to be sec^4(x) tan^4(x) ?

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

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5. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

I'm so sorry but that way of showing integration by parts confuses me so much, without certain intermediate steps and using u and v I'm pretty lost. Take the first example you did:

So we work in a counter clockwise direction starting at the bottom left. OK. so the problem splits to lnx+1/sqrt(x) then from there a bunch of hidden intermediate steps happen and it's kinda difficult to follow how this was done out when I wasn't taught this way =/

6. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Ok I do usually show less in the first picture, which makes the direction of travel a bit clearer. Which is, roughly...

Integration by parts (however you do it) is about working backwards through the product rule.

So you want to see the integrand as the result of a product rule differentiation.

But the result of a product rule differentiation is one product plus another. The bottom row here...

You have just the one product...

Nonetheless, you can fill out the inevitable consequences of putting your product in that place...

(only inevitable once you have chosen between legs crossed and uncrossed, i.e. which part to integrate immediately and which not)

... and then cancel those consequences out, like this, so that the bottom row is again equal to what you wanted to integrate...

And then you can complete the top row so that it differentiates to give the bottom...

... which is equal to your original product (the integrand).

That's exactly what the u and v business is doing, too, of course. It's a handy formula if you know how to use it, but if you can't square it with the above you probably haven't grasped how it works, and will get hung up on the notation (the u and v).

7. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by goku900
) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx

2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

3) 0 to pi ∫ e^(cosx)sin(2x) dx

4) 4 to 9∫ ln(x)/sqrt(x) dx

5) Derive the basic Integral

∫csc(x)dx=ln[cot(x)-csc(x)]+C

Help on any of these would be appreciated.
\displaystyle \displaystyle \begin{align*} \int_0^{\frac{\pi}{4}}{\sec^4{(x)}\tan^4{(x)}\,dx} &= \int_0^{\frac{\pi}{4}}{\sec^2{(x)}\sec^2{(x)}\tan^ 4{(x)}\,dx} \\ &= \int_0^{\frac{\pi}{4}}{\sec^2{(x)}\left[ 1 + \tan^2{(x)} \right] \tan^4{(x)}\,dx} \\ &= \int_0^{\frac{\pi}{4}}{\sec^2{(x)}\left[ \tan^4{(x)} + \tan^6{(x)}\right] dx} \end{align*}

Now let $\displaystyle \displaystyle u = \tan{(x)} \implies du = \sec^2{(x)}\,dx$ and the integral becomes

\displaystyle \displaystyle \begin{align*} \int_0^{\frac{\pi}{4}}{\sec^2{(x)}\left[ \tan^4{(x)} + \tan^6{(x)} \right] dx} &= \int_0^{1}{u^4 + u^6\,du} \\ &= \left[ \frac{u^5}{5} + \frac{u^7}{7} \right] _0 ^1 \\ &= \frac{1}{5} + \frac{1}{7} \\ &= \frac{12}{35} \end{align*}

8. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by goku900
) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx

2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

3) 0 to pi ∫ e^(cosx)sin(2x) dx

4) 4 to 9∫ ln(x)/sqrt(x) dx

5) Derive the basic Integral

∫csc(x)dx=ln[cot(x)-csc(x)]+C

Help on any of these would be appreciated.
$\displaystyle \displaystyle \int{\frac{\ln{(x)}}{\sqrt{x}}\,dx}$

Use integration by parts with $\displaystyle u = \ln{(x)} \implies du = \frac{1}{x}\,dx$ and $\displaystyle \displaystyle dv = \frac{1}{\sqrt{x}}\,dx \implies v = 2\sqrt{x}$ and the integral becomes

\displaystyle \displaystyle \begin{align*} \int{\frac{\ln{(x)}}{\sqrt{x}}\,dx} &= 2\sqrt{x}\ln{(x)} - \int{ \frac{2\sqrt{(x)}}{x}\,dx } \\ &= 2\sqrt{x}\ln{(x)} - \int{\frac{2}{\sqrt{x}}\,dx} \\ &= 2\sqrt{x}\ln{(x)} - 4\sqrt{x} + C \end{align*}

9. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by goku900
) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx

2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

3) 0 to pi ∫ e^(cosx)sin(2x) dx

4) 4 to 9∫ ln(x)/sqrt(x) dx

5) Derive the basic Integral

∫csc(x)dx=ln[cot(x)-csc(x)]+C

Help on any of these would be appreciated.
\displaystyle \displaystyle \begin{align*} \int{\csc{(x)}\,dx} &= \int{\frac{1}{\sin{(x)}}\,dx} \\ &= \int{\frac{\sin{(x)}}{\sin^2{(x)}}\,dx} \\ &= -\int{\frac{-\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \\ &= -\int{\frac{1}{1 - u^2}\,du} \textrm{ after making the substitution } u = \cos{(x)} \implies du = -\sin{(x)}\,dx \\ &= -\int{\frac{1}{(1-u)(1+u)}\,du} \\ &= -\int{\frac{\frac{1}{2}}{1-u} + \frac{\frac{1}{2}}{1+u}\,du} \\ &= -\left[ -\frac{1}{2}\ln{|1-u|} + \frac{1}{2}\ln{|1+u|} \right] + C \\ &= \frac{1}{2}\ln{\left| \frac{1-u}{1+u} \right|} + C \\ &= \frac{1}{2}\ln{\left| \frac{1 - \cos{(x)}}{1 + \cos{(x)}} \right|} + C \\ &= \frac{1}{2} \ln{\left| \frac{\left[ 1 - \cos{(x)} \right] ^2 }{\left[ 1 + \cos{(x)} \right] \left[ 1 - \cos{(x)} \right] } \right| } + C \end{align*}

\displaystyle \displaystyle \begin{align*} &= \frac{1}{2}\ln{ \left| \frac{ \left[ 1 - \cos{(x)} \right] ^2 }{ 1 - \cos^2{(x)} } \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{ \left[ 1 - \cos{(x)} \right] ^2 }{ \sin^2{(x)} } \right| } + C \\ &= \ln{ \left\{ \left| \frac{ \left[ 1 - \cos{(x)} \right] ^2 }{ \sin^2{(x)} } \right| ^{\frac{1}{2}} \right\} } + C \\ &= \ln{ \left| \frac{1 - \cos{(x)}}{\sin{(x)}} \right| } + C \\ &= \ln{ \left| \frac{1}{\sin{(x)}} - \frac{\cos{(x)}}{\sin{(x)}} \right| } + C \\ &= \ln{ \left| \csc{(x)} - \cot{(x)} \right| } + C \end{align*}

10. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

The csc(x) integration is tricky, thanks a bunch tho. I have questions

In the third line you make the sin in the numerator negative and also pull a negative out of the integral. Why do you need to do this?

Also how would anyone know to break up the 1/(1-u)(1+u) into 1/2 in the numerator...?!?

Yeah that negative sign in the beggining is really confusing to me, after you get the first solution with the +C, explain how that breaks up into that 1-u/1+u) term

11. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Why did you change the boundaries to 0 to 1 on this problem

lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

12. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by goku900
The csc(x) integration is tricky, thanks a bunch tho. I have questions

In the third line you make the sin in the numerator negative and also pull a negative out of the integral. Why do you need to do this?

Also how would anyone know to break up the 1/(1-u)(1+u) into 1/2 in the numerator...?!?

Yeah that negative sign in the beggining is really confusing to me, after you get the first solution with the +C, explain how that breaks up into that 1-u/1+u) term
I made it negative because when you make the substitution u = cos(x), you need to note that its derivative is du/dx = -sin(x), or du = -sin(x) dx. In order to turn it into the du integral, we need to replace -sin(x) dx with du.

Originally Posted by goku900
Why did you change the boundaries to 0 to 1 on this problem

lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx
Remember that u = tan(x). I changed the bounds because when x = 0, u = tan(0) = 0 and when x = pi/4, u = tan(pi/4) = 1. By doing this, we avoid having to convert back to a function of x after integration, we can substitute the new u bounds straight away.

13. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

The integral of $\displaystyle \csc{x}$ has a "trick" shortcut method:

$\displaystyle \int\csc{x}\,dx =$

$\displaystyle \int\csc{x}\left(\frac{\csc{x} - \cot{x}}{\csc{x}-\cot{x}}\right)\,dx =$

$\displaystyle \int\frac{\csc^2{x}-\csc{x}\cot{x}}{\csc{x}-\cot{x}}\,dx =$

and since $\displaystyle \frac{d}{dx}\csc{x}=-\cot{x}\csc{x}$ and $\displaystyle \frac{d}{dx}\cot{x}=-\csc^2{x}$, the numerator is the derivative of the denominator, so

$\displaystyle \int\csc{x}\,dx =\ln{|\csc{x}-\cot{x}|}+C$

- Hollywood

14. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by goku900
) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx

2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

3) 0 to pi ∫ e^(cosx)sin(2x) dx

4) 4 to 9∫ ln(x)/sqrt(x) dx

5) Derive the basic Integral

∫csc(x)dx=ln[cot(x)-csc(x)]+C

Help on any of these would be appreciated.

$\displaystyle \displaystyle\int_2^4\frac{\sqrt{9x^2 - 25}}{x^3}dx$

let $\displaystyle \space$ $\displaystyle u = 3x$; $\displaystyle \space$ $\displaystyle x = \displaystyle\frac{u}{3}$; $\displaystyle \space$ $\displaystyle x^3 = \displaystyle\frac{u^3}{27}$

$\displaystyle du = 3dx$; $\displaystyle \space$ $\displaystyle dx = \displaystyle\frac{1}{3}du$

$\displaystyle \sqrt {9x^2 - 25} = \sqrt{u^2 - 25}$

$\displaystyle \displaystyle\int_2^4\frac{\sqrt{9x^2 - 25}}{x^3}dx = \displaystyle\int_6^{12}\frac{\displaystyle\frac{1 }{3}\sqrt{u^2 - 25}}{\displaystyle\frac{1}{27}u^3}du = 9\displaystyle\int_6^{12}\frac{\sqrt{u^2 - 25}}{u^3}du$

Let $\displaystyle \space$ $\displaystyle u = \displaystyle\frac{5}{\sin {\theta}} = 5\csc{\theta}$ $\displaystyle \space$; $\displaystyle u^3 = 125\csc^3{\theta}$

$\displaystyle \sqrt{u^2 - 25} = 5\cot{\theta}$; $\displaystyle \space$ $\displaystyle du = -5\csc{\theta}\cot{\theta}d{\theta}$

$\displaystyle 9\displaystyle\int_6^{12}\frac{\sqrt{u^2 - 25}}{u^3}du = -\displaystyle\frac{9}{5}\int_{\sin^{-1}{\frac{5}{6}}}^{\sin^{-1}{\frac{5}{12}}}\displaystyle\frac{\cot^2{\theta} }{\csc^2{\theta}}d{\theta} = -\displaystyle\frac{9}{5}\int_{\sin^{-1}{\frac{5}{6}}}^{\sin^{-1}{\frac{5}{12}}}\cos^2{\theta}d{\theta}$

$\displaystyle -\displaystyle\frac{9}{5}\int\cos^2{\theta}d{\theta } = -\displaystyle\frac{9}{5}\displaystyle(\frac{\theta }{2} + \displaystyle\frac{\sin 2{\theta}}{4}) + C = \displaystyle-\frac{9}{10}\tan^{-1}{\frac{5}{\sqrt{9x^2 - 25}} - \displaystyle\frac{\sqrt{9x^2 - 25}}{2x^2} + C$

From there you should be able to evaluate the definite integral.

Someone check me on this to see if I am correct. Thanks

15. Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Originally Posted by jpritch422
$\displaystyle \displaystyle\int_2^4\frac{\sqrt{9x^2 - 25}}{x^3}dx$

let $\displaystyle \space$ $\displaystyle u = 3x$; $\displaystyle \space$ $\displaystyle x = \displaystyle\frac{u}{3}$; $\displaystyle \space$ $\displaystyle x^3 = \displaystyle\frac{u^3}{27}$

$\displaystyle du = 3dx$; $\displaystyle \space$ $\displaystyle dx = \displaystyle\frac{1}{3}du$

$\displaystyle \sqrt {9x^2 - 25} = \sqrt{u^2 - 25}$

$\displaystyle \displaystyle\int_2^4\frac{\sqrt{9x^2 - 25}}{x^3}dx = \displaystyle\int_6^{12}\frac{\displaystyle\frac{1 }{3}\sqrt{u^2 - 25}}{\displaystyle\frac{1}{27}u^3}du = 9\displaystyle\int_6^{12}\frac{\sqrt{u^2 - 25}}{u^3}du$

Let $\displaystyle \space$ $\displaystyle u = \displaystyle\frac{5}{\sin {\theta}} = 5\csc{\theta}$ $\displaystyle \space$; $\displaystyle u^3 = 125\csc^3{\theta}$

$\displaystyle \sqrt{u^2 - 25} = 5\cot{\theta}$; $\displaystyle \space$ $\displaystyle du = -5\csc{\theta}\cot{\theta}d{\theta}$

$\displaystyle 9\displaystyle\int_6^{12}\frac{\sqrt{u^2 - 25}}{u^3}du = -\displaystyle\frac{9}{5}\int_{\sin^{-1}{\frac{5}{6}}}^{\sin^{-1}{\frac{5}{12}}}\displaystyle\frac{\cot^2{\theta} }{\csc^2{\theta}}d{\theta} = -\displaystyle\frac{9}{5}\int_{\sin^{-1}{\frac{5}{6}}}^{\sin^{-1}{\frac{5}{12}}}\cos^2{\theta}d{\theta}$

$\displaystyle -\displaystyle\frac{9}{5}\int\cos^2{\theta}d{\theta } = -\displaystyle\frac{9}{5}\displaystyle(\frac{\theta }{2} + \displaystyle\frac{\sin 2{\theta}}{4}) + C = \displaystyle-\frac{9}{10}\tan^{-1}{\frac{5}{\sqrt{9x^2 - 25}} - \displaystyle\frac{\sqrt{9x^2 - 25}}{2x^2} + C$

From there you should be able to evaluate the definite integral.

Someone check me on this to see if I am correct. Thanks
I solved using trig substitution and got .5148. Using the boundaries 2 to 4

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