Question

I need help with integrals.

Answer 1

Ooo the fun stuff.
Have a specific problem you're working on? :)

Answer 2

((sinx-cosx)/(sinx+cosx))^4 dx
Limit- from 0 to pi/4
Thank you for answering.

Answer 3

Hmm this one looks tricky :D thinking

Answer 4

Any solution👀

Answer 5

?

Answer 6

You posted the problem correctly?
The entire thing is 4th power?

It works out much nicer if it's just denominator 4th power,
so just asking :)

Answer 7

I post the problem correctly.👍

Answer 8

Hmmmm I'm not sure then :c

Answer 9

We can use x=pi/4-t

Answer 10

Oke then never mind.😕

Answer 11

?

Answer 12

Also integral of sin^(6x)sin^(3x)dx.
And integral of arccossqrt(x/x+1)dx.
I am waiting for answers. .

Answer 13

"I am waiting for answers" - There's the problem. This isn't an answers site.

That said, I'd be happy to give a nudge or two in the right direction.

First integral:
\[\int_0^{\pi/4}\frac{\sin x-\cos x}{(\sin x+\cos x)^4}\,\mathrm{d}x\]Hint: If \(u=\sin x+\cos x\), then \(\mathrm{d}u=(\cos x-\sin x)\,\mathrm{d}x\), and so this integral is equivalent to
\[-\int_1^{\sqrt2}\frac{\mathrm{d}u}{u^4}\]Alternatively, you suggest substituting \(x=\dfrac{\pi}{4}-t\), which is an excellent idea, as this gives \(\mathrm{d}x=-\mathrm{d}t\) and with our handy trigonometric identities the integral becomes
\[\begin{align*}-\int_{\pi/4}^0\frac{\sin\left(\dfrac{\pi}{4}-t\right)-\cos\left(\dfrac{\pi}{4}-t\right)}{\left(\sin\left(\dfrac{\pi}{4}-t\right)+\cos\left(\dfrac{\pi}{4}-t\right)\right)^4}\,\mathrm{d}t&=\int_0^{\pi/4}\frac{-\sqrt2\sin t}{\sqrt2\cos t}\,\mathrm{d}t\\[1ex]&=-\int_0^{\pi/4}\tan t\,\mathrm{d}t\end{align*}\]
Second integral: Not sure if you mean\(\sin^6x\sin^3x\) or \(\sin6x\sin3x\), or some combination of the two. Clarification would be appreciated.

Third integral:
\[\int\arccos\left(\sqrt\frac{x}{x+1}\right)\,\mathrm{d}x\]
I bet this admits a neat substitution somewhere along the line, but standalone inverse trig functions in this form often can be simplified via integration by parts. Setting
\[\begin{matrix}u=\arccos\left(\sqrt\frac{x}{x+1}\right)&&&\mathrm{d}v=\mathrm{d}x\\[1ex]
\mathrm{d}u=-\dfrac{1}{2x}\sqrt{\dfrac{x}{(x+1)^2}}\,\mathrm{d}x&&&v=x\end{matrix}\]so the integral becomes
\[x\arccos\left(\sqrt\frac{x}{x+1}\right)+\frac{1}{2}\int\sqrt\frac{x}{(x+1)^2}\,\mathrm{d}x\]which looks easier to work with.

Answer 14

Sin^2(6x)sin^2 (3x)dx

Answer 15

I do not understand the third one .. we don't have x for dv .

Answer 16

And the first one is all integral of 4th power not just denominator

Answer 17

My mistake, the first one should be
\[\begin{align*}-\int_{\pi/4}^0\left(\frac{\sin\left(\dfrac{\pi}{4}-t\right)-\cos\left(\dfrac{\pi}{4}-t\right)}{\sin\left(\dfrac{\pi}{4}-t\right)+\cos\left(\dfrac{\pi}{4}-t\right)}\right)^4\,\mathrm{d}t&=\int_0^{\pi/4}\left(\frac{-\sqrt2\sin t}{\sqrt2\cos t}\,\mathrm{d}t\right)^4\\[1ex]&=-\int_0^{\pi/4}\tan^4 t\,\mathrm{d}t\end{align*}\]Noting that \(\tan^4t=\tan^2t(\sec^2t-1)\), you can expand to obtain
\[-\int_0^{\pi/4}(\tan^2t\sec^2t-\tan^2t)\,\mathrm{d}t=\int_0^{\pi/4}(\sec^2t(1-\tan^2t)-1)\,\mathrm{d}t\]

For the second integral, one thing you can do is to write
\[\int\left(\sin(6x)\sin(3x)\right)^2\,\mathrm{d}x=\int\left(\frac{\cos(3x)-\cos(9x)}{2}\right)^2\,\mathrm{d}x\]Expanding, you get
\[\frac{1}{4}\int(\cos^2(3x)-2\cos(3x)\cos(9x)+\cos^2(9x))\,\mathrm{d}x\]which is equivalent to
\[\frac{1}{4}\int\left(\cos^2(3x)-2\left(\frac{\cos(6x)+\cos(12x)}{2}\right)+\cos^2(9x)\right)\,\mathrm{d}x\]The same identity lets you rewrite this as
\[\frac{1}{4}\int\left(\frac{1+\cos6x}{2}-(\cos(6x)+\cos(12x))+\frac{1+\cos(18x)}{2}\right)\,\mathrm{d}x\]

For the last one, are you not familiar with integration by parts? Or just having trouble following the procedure above?

Answer 18

Is Solution of second integral x/4-sin6x/48-sin12/48+sin18/144

Answer 19

Last one I clearly unterstand :D